Oh? Just what is the magical term in them that prevents
(x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true?

It turns out not to be magic, but reality, that interferes
with the application of the galilean transforms to the gen-
eralized coordinate form(s) of Maxwell: there are no coordi-
nates to transform!

When True Believer crackpots are shown the simple
demonstration that the galilean transform on 
generalized cartesian coordinates is invariant,
their first defense is usually an incredibly stupid
"x0'=x0, because the coordinate of a circle center,
or point of emission, etc, is a constant and can't
be transformed."

The last defense is "but Maxwell's equations are not
invariant under that coordinate transform."  When
asked just what magic occurs in Maxwell that would
prevent the simple algebra 

    (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0) 

from working, and when asked them for a demonstration,
they will never do so, however many hundreds of
times their defense is asserted.

The reason may help you understand part of Einstein's
1905 paper in which he gave us his absurd Special
Relativity derivation:

THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.

Einstein gave the electric force vector as E=(X,Y,Z)
and the magnetic force vector as B=(L,M,N), where the
force components in the direction of the x axis are
X and L,  Y and M are in the y direction, Z and N in
the z direction.

Those values are not, however, coordinates, but values
very much like acceleration values.

BTW, the current fad is that E and B are 'fields', having
been 'force fields' for a while, after being 'forces'.

So, when Einstein says he is applying his coordinate
transforms to the Maxwell form he presented, he is 
either delusive or lying.

(a) there are no coordinates in the transform equations
    he gives us for the Maxwell transforms, where
    B=beta=1/sqrt(1-(v/c)^2):
     
    X'=X.                L'=L.
    Y'=B(Y-(v/c)N).      M'=B(M+(v/c)Z).
    Z'=B(Z+(v/c)M).      N'=B(N-(v/c)Y).

    X is in the same direction as x, but is not a coordinate.
    Ditto for L. They are not locations, coordinates on the
    x-axis, but force magnitudes in that direction.

    Similarly for Y and M and y, Z and N and z.

(b) the v of the "coordinate transforms" is in Maxwell
    before any transform is imposed; Einstein's transform
    v is the velocity of a coordinate axis, not the velocity
    of a particle, which is what was in the equation before
    he touched it.

(c) if they were honest Einsteinian transforms, they'd be
    incompetent. The direction of the particle's movement is 
    x, which means it is X and L that are supposed to be 
    transformed, not Y and M, and Z and N. And when SR does 
    transform more than one axis, each axis has its own 
    velocity term;  using the v along the x-axis as the v 
    for a y-axis and z-axis transform is thus trebly absurd: 
    the axes perpendicular to the motion are not changed
    according to SR, the v used is not their v, and the v 
    is not a transform velocity anyway.

(d) as everyone knows, the effect of E and B are on the
    particle's velocity, which is a speed in a particular
    direction.  Both the speed and direction are changed
    by E and B, but v - the speed - is a constant in SR.

As absurd as are the previously demonstrated Einsteinian 
blunders, this one transcends error and is an incredible 
example of True Believer delusion propagating over decades.

The components of E and B do differ from point to point,
and in the variations that are not coordinate free,
they are subject to the usual invariant galilean trans-
formation when put in the generalized coordinate form.

-------------------------------------------------------------

The SR crackpots don't know what coordinates are. The 
various things they call coordinates include coordin-
nates, but also include a variety of other quantities.

------------------------------------------------------

1.   One may express coordinates in a one-axis-at-a-time
     manner [like x^2+y^2=r^2] but it is the use of vector
     notation that shows us what is going on. In vector
     notation the triplet x,y,z [or x1,x2,x3, whatever]
     represents the three spatial coordinates, but there
     are so-called basis vectors that underlie them. Those
     may be called i,j,k. Thus, what we normally treat as
     x,y,z is a set of three numbers TIMES a basis vector
     each. 

2.   These e*i, f*j, g*k products can have a lot of meanings.
     
     If e, f, j are distances from the origin of i,j,k then
     e*i, f*j, g*k are coordinates: distances in the directions
     of i,j,k respectively, from their origin. That makes the 
     triplet a coordinate vector that we describe as being an
     x,y,z triplet; perhaps X=(x,y,z).

     The e*i, f*j, g*k products could be directions; take any
     of the other vectors described above or below and divide the
     e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. 
     That gives us a vector of length=1.0, the e,f,g values of 
     which show us the direction of the original vector. That 
     makes the triplet a direction vector that we describe as 
     being an x,y,z triplet; perhaps D=(x,y,z).

     The e*i, f*j, g*k products could be velocities; take any
     of the unit direction vectors described above and multiply
     by a given speed, perhaps v. That gives a vector of length 
     v in the direction specified. That makes the triplet a 
     velocity vector that we describe as being an x,y,z triplet; 
     perhaps V=(x,y,z). Each of the three values, e,f,g, is the 
     velocity in the direction of i,j,k respectively.

     The e*i, f*j, g*k products could be accelerations; take any
     of the unit direction vectors described above and multiply
     by a given acceleration, perhaps a. That gives a vector of 
     length a in the direction specified. That makes the triplet 
     an acceleration vector that we describe as being an x,y,z 
     triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, 
     is the acceleration in the direction of i,j,k respectively.
     
     The e*i, f*j, g*k products could be forces (much like accel-
     erations); take any of the unit direction vectors described 
     above and multiply by a given force, perhaps E or B. That 
     gives a vector of length E or B in the direction specified. 
     That makes the triplet a force vector that we describe as 
     being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each 
     of the three values, e,f,g, is the force in the direction of 
     i,j,k respectively.


Einstein's - and Maxwell's - E and B are 
not coordinate vectors.

============================================================

There is another variety of intellectual befuddlement that 
misinforms the idea that Maxwell isn't invariant under the
galilean transform: confusions about velocities.

Velocities With Respect to Coordinate Systems.
-----------------------------------------------
Aaron Bergman supplied the background in a post to a sci.physics.* 
newsgroup:
===============================================================

Imagine two wires next to each other with a current I in each. 
Now, according to simple E&M, each current generates a magnetic 
field and this causes either a repulsion or attraction between 
the wires due to the interaction of the magnetic field and the 
current. Let's just use the case where the currents are parallel. 
Now, suppose you are running at the speed of the current between 
the wires. If you simply use a galilean transform, each wire, 
having an equal number of protons and electrons is neutral. So,
in this frame, there is no force between the wires. But this is a
contradiction.

================================================================

First of all, the invariance of the galilean transform (x'-x.c')
=(x-x.c),  insures that it is an error to imagine there is any
difference between the data and law in one frame and in another;
the usual, convenient rest frame is the best frame and only frame 
required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but
(x'-x.c')=(x-x.c).]

Second, given that you decide unnecessarily to adapt a law to 
a moving frame, don't confuse coordinate systems with meaningful 
physical objects, like the velocity relative to a coordinate 
system instead of relative to a physical body or field.  

In other words, what does current velocity with respect to a 
coordinate system have to do with physics? 

Nothing. Certainly not anything in the example Bergman gave.

What is relevant is not current velocity with respect to a 
coordinate system, but current velocity with respect to wires
and/or a medium.  The velocity of an imaginary coordinate sys-
tem has absolutely nothing to do with meaningful physical vel-
ocity. You can - if you are insightful enough and don't violate
item (e) - identify a coordinate system and a relevant physical
object, but where some v term in the pre-transformed law is
in use, don't confuse it with the velocity of the coordinate 
transform.


Velocities With Respect to ... What?
-----------------------------------------------
Albert Einstein opened his 1905 paper on Special Relativity
with this ancient incompetency:
===============================================================

The equations of the day had a velocity term that was taken
as meaning that moving a magnet near a conductor would create
a current in the conductor, but moving a conductor near a
wire would not.  This was belied by fact, of course.

The important velocity quantity is the velocity of the 
magnet and conductor with respect to each other, not to
some absolute coordinate frame (as far as we know) and
not to an arbitrary coordinate system.

One possible cause was the idea: "but the equation says the magnet
must be moving wrt the coordinate system" or "... the absolute
rest frame". 

There not being anything in the equation(s) to say either of
those, it is amazing that folk will still insist the velocity
term has nothing to do with velocity of the two bodies wrt
each other.
-----------------------------------------------------------