In the rest frame nothing moves (i.e. no forces), so in the boosted frame the same must hold (besides the objects moving according to the boost of course). Our material is essentially a collection of internal magnetic dipoles, so we are reduced to the scenario of a single magnetic moment $\mu$ sitting with a stationary charge. When you boost, a magnetic field is produced from this charge, and hence $\mu$ should realign to this $B$. But we know that this $\mu$ must not actually move, so what's the deal? Well because we boosted, $\mu$ itself changed! It's motion will counteract the intended realignment, resulting in no movement.
Elaboration:
We have to be careful on what we mean by a 'magnetic dipole'... this has caused great confusion in the past, and involved the concept of "hidden EM angular momentum" and leads to the Mansuripur paradox (which is precisely analogous to our situation!). For more information, see here:
http://www.physics.princeton.edu/~mcdonald/examples/mansuripur.pdf
http://ajp.aapt.org/resource/1/ajpias/v57/i2/p171_s1?isAuthorized=no
So we will just assume we are working with the right definition of a magnetic dipole, let's say not the current-loop definition, so that we can avoid this hidden EM momentum issue and other dipole-torque issues.
We start with a dipole moment $\mu$ in a static electric field (produced by some charge $q$), experiencing no force/torque. When we boost (speed $v$ in some convenient direction, also nonrelativistic so I can throw away my $\gamma$'s), a magnetic field by the now-moving charge is produced, and so there is a resulting torque $N_1=\mu\times B$ which tends to align $\mu$ with $B$. But by invariance, this $\mu$ cannot actually move, so we are missing something. What we are missing is the fact that $\mu$ was boosted and hence has an electric dipole moment $p=v\times \mu/c$. There is then a torque $N_2=p\times E$ on this dipole moment from the (boosted) electric field $E$.
Now the total torque in any reference frame is $N_\text{tot}=p\times E+\mu\times B + \frac{1}{c}v\times(p\times B)- \frac{1}{c}v\times(\mu\times E)$, which evaluates to zero here (before and after the Lorentz transformations).