Does a magnet moving in a uniform electric field experience torque? Assume a uniform electric field of $E_y$ along $y$ in the lab frame of reference $(x,y,z,t)$. A simple magnet bar is set in motion at $v$ along $x$ in this electric field so that the alignment of the bar is along $x$ too. Please, tell me what happens from the viewpoint of each of the observers, one located on the magnet $(x^\prime,y^\prime,z^\prime,t^\prime)$ and the other at rest with respect to the electric field (the lab observer). 
My own guess at the answer is that the magnet would rotate 90 degrees in plane $xz$ as seen by the lab observer. Because, from the perspective of the observer located on the magnet, the electric field moves at $v$, and thus, applying the Lorentz transformation for fields, a magnetic field of ${B^\prime}_{z^\prime}=-\gamma v E_y/c^2$ is produced along $z^\prime$. This magnetic field tends to produce a torque on the magnet and rotate it in plane $xz$. However, how does the lab observer justify this rotation?
 A: The easiest way to reconcile the perspective of the magnet observer and the lab observer is using the covariant formulation of all of the relevant quantities. In this case the important quantity is the magnetization-polarization tensor $\mathbf M$ (I will use bold for covariant quantities, the $(+\, -\, -\, - )$ signature, and units where c=1). The components of $\mathbf M$ are $$ \mathbf{M}^{\mu\nu}=
\left(
\begin{array}{cccc}
 0 & P_x & P_y & P_z \\
 -P_x & 0 & -M_z & M_y \\
 -P_y & M_z & 0 & -M_x \\
 -P_z & -M_y & M_x & 0 \\
\end{array}
\right)$$ where $M_i$ and $P_i$ are the $i$th components of the ordinary magnetization and polarization three-vectors. Introducing the the four gradient $\mathbf{\partial}$ and the four-current $\mathbf J$, we can write the bound four-current as $\mathbf J^{\nu} = \mathbf{\partial}_{\mu} \mathbf M^{\mu\nu}$
This immediately shows us that what is a pure magnetization in one frame may be both a magnetization and a polarization in another frame. Intuitively this gives us an understanding that we can generally expect for there to be non-zero bound charges on the surface of a bar magnet in a frame where the magnet is moving. In other words, the bound current in the magnet frame can transform into a bound current and bound charge in another frame.
Now, to show that the perspective of the magnet observer and the perspective of the lab observer are consistent, it is sufficient to write the Lorentz four-force density $\mathbf f$ on the magnet in a manifestly covariant way. We introduce the electromagnetic field tensor $\mathbf F$. Then we can write $$\mathbf f_{\mu} = \mathbf F_{\mu\nu} \mathbf J^{\nu} = \mathbf F_{\mu\nu} \mathbf{\partial}_{\xi} \mathbf M^{\xi\nu}$$
This expression is manifestly covariant and can be used in both frames to calculate the four-force density on the magnet. Therefore we have shown that the perspectives of both observers are guaranteed to be consistent.
For the specific scenario in the question the results are fairly interesting. If we assume a cubic shaped magnet with the magnetization along $x'$ then we can write $$\mathbf M'^{\mu\nu}=\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & -M \ \Pi (x',y',z') \\
 0 & 0 & M \ \Pi (x',y',z') & 0 \\
\end{array}
\right)$$ where $\Pi$ is the multidimensional Heaviside pi function which is 1 if all of the arguments are between -1/2 and 1/2 and is 0 otherwise. 
The partial derivative of $\Pi$ is $$\partial_x \Pi(x,y,z) =(2 \delta (2 x+1)-2 \delta (2 x-1)) \Pi (y,z)$$ where $\delta$ is the Dirac delta function. Thus $\partial_x \Pi(x,y,z)$ is a function which represents a pair of opposite surface charges or surface currents on the $+x$ and $-x$ surface of the cube with it being positive on the $-x$ surface and negative on the $+x$ surface. Because this shows up a lot, I will define $s(x)=\partial_x \Pi(x,y,z)$ for the $+x$ and $-x$ surfaces, and similarly for the other directions.
The bound currents are thus $\mathbf J'^{\nu}=\partial_{\mu} \mathbf M'^{\mu\nu} = \left( 0,0,M \ s(z'), M \ s(y') \right)$ as expected. The Lorentz force density is then $\mathbf f'_{\mu}=\mathbf F'_{\mu\nu} \mathbf J'^{\nu} = \left( - \gamma \ E_y \ M \ s(z'), - \gamma \ E_y \ M \ v \ s(z'), 0, 0 \right)$. So there is a non-zero force in the $x'$ direction on the $z'$ surfaces of the magnet.
Interestingly, transforming to the lab frame we get $$\mathbf f_{\mu} = \Lambda_{\mu}^{\mu'} \mathbf f'_{\mu'} = \mathbf F_{\mu\nu} \mathbf{\partial}_{\xi} \mathbf M^{\xi\nu} $$ $$\mathbf f_{\mu} = \left( -E_y \ M \ s(z'), 0, 0, 0 \right) $$
So, the magnitude and location of the forces in the primed frame are such that they disappear when transforming to the unprimed frame. In other words, the OP's analysis of the forces is correct, there is a non-zero pair of surface forces on the $z'$ surfaces in the magnet frame, and no surface forces in the lab frame! 
Note however, that the rotation occurs in both frames. In the magnet frame it is explained by the forces. In the lab frame although there is no spatial component to the four force there is a time component. This time component increases the energy of one side and decreases the energy of the other side. Since the velocity is the momentum divided by the energy an increase in energy without an increase in momentum results in a reduction in velocity. So the velocity of one side increases and the velocity of the other decreases without any change in momentum and therefore no force. 
Therefore, this scenario is not in conflict with either relativity or electromagnetism. In fact, transforming the forces correctly leads to exactly the correct outcome in all frames, thus the Lorentz force is fully compatible with relativity as shown above. The confusion in the OP arises from a failure to correctly transform the forces between frames, to recognize that they are expected to vanish, and to account for the kinematic effects of the energy.
A: Lab frame
Let's say the magnet is a wire loop with current, and at first we orient the loop like this:  O 
The loop moves to the right and there is an electric field pointing into the screen.
Special relativity tells us that electrons spend different time at the upper part compared to the lower part. So the loop is an electric dipole that experiences a torque in the electric field.
Now we reorient the loop so that it looks like this:  |  
Special relativity tells us that the loop is not an electric dipole now, so it does not experience a torque in the electric field.
Magnet frame
First we have the loop oriented like this: O
There is a negatively charged plate that creates the aforementioned electric field, the plate is moving in this frame.
There is an attractive magnetic force between electrons that are moving to the same direction in the loop and in the plate. And a repulsive magnetic force between electrons that are moving into opposite directions in the loop and in the plate. Those forces cause a torque on the loop.  
Now we reorient the loop as before. This loop: O becomes this loop: | 
Now electrons in the loop move in yz plane, while electrons in the plate move along the x-coordinate. There are no magnetic forces between electrons in the loop and electrons in the plate. So there is no torque.
A: @Dale :

Note however, that the rotation occurs in both frames. In the magnet frame it is explained by the forces. In the lab frame although there is no spatial component to the four force there is a time component. This time component increases the energy of one side and decreases the energy of the other side.

Contrary to your claim, I want to show you that there are also spatial force components exerting on the magnet from the viewpoint of the observer at rest in the lab frame. This observer claims that the magnet moves along $x$, thus we have a non-zero $\partial B/\partial t$, which produces a circulating electric field around the magnet in plane $xz$.
As these field lines pass through the charges of the capacitor that produce the uniform electric field, they tend to move the capacitor charges along $z$. Remember that the direction of these circulating electric fields is different for the magnet poles; one is clockwise and the other counter-clockwise, that is. On the other hand, for a specific magnetic pole, the direction of the forces exerted on the positively and negatively charged plates, due to the circulating $E$-fields, is the same.
Therefore, If the circulating $E$-field around pole $S$ of the magnet exerts a force along $+z$ on the plates, this force would be along $-z$ for the circulating $E$-field around the $N$ pole. The reactions of these forces exerting on the plates, appear on the magnet and make it rotate in plan $xz$. Therefore, it is evident that there are also some spatial force components on the magnet from the viewpoint of the lab observer in contrast to your claim, and therefore:

Intuitive is a matter of taste and experience.

This is not always the case!
