According to the Lorentz transformation formula for the electric field, we have:

$$ \mathbf E_\perp' = \gamma(\mathbf E_\perp+\mathbf v\times \mathbf B_\perp) $$

Since $\mathbf B=0$ in the rest frame, in the moving frame $\mathbf E$ becomes stronger by a factor of $\gamma$. We expect these effects to cancel out. The size of the magnetic field generated is $|B_\perp| = \gamma v|E_\perp|/c^2$, by a similar Lorentz transformation law.

Lorentz force on the moving charge will be $qv|B_\perp|$, since the field is perpendicular to the direction of motion. So the overall force due to magnetic field is $\gamma |E_\perp| qv^2/c^2$. Meanwhile, the magnitude of the electric force on the charge is the direction perpendicular to motion is $\gamma |E_\perp| q$. Since these are pointing in opposite directions, the net attractive force, perpendicular to the direction of motion, is:

$$ \gamma |E_\perp|q(1-\frac{v^2}{c^2}) = \frac{1}{\sqrt{1-v^2/c^2}}|E_\perp|q(1-v^2/c^2) = |E_\perp|q\sqrt{1-v^2/c^2} $$

This is not a complete cancellation, the perpendicular electromagnetic force is still reduced in the moving frame. The only chance now is that maybe in the moving frame, our insulating stick applies a force that isn't parallel to itself.

In the resting frame, the stick transmits a force $F$ between the two charges. The definition of force is $dp/dt$: the rate of flow of momentum. In the resting frame, both charges have 0 momentum at all times. So the net momentum change is 0. Some momentum is flowing through the electric field between the charges at a rate of $F$, and an equal and opposite quantity of momentum is flowing through the insulating stick, also at a rate of $F$.

Let's consider the change in force applied by the stick after our Lorentz transformation. In the rest frame, we have time coordinate $t$. In the moving frame, we have time coordinate $t' = \gamma(t-vx/c^2)$. At rest, $x=0$ at all times, so $t' = \gamma t$. This means that a small interval of time $dt$ in the rest frame corresponds to a slightly longer interval of time in the moving frame, $dt'=\gamma dt$. The momentum transferred in a time $dt$ is $dp = (Fdt\cos\alpha, Fdt\sin\alpha)$ in the rest frame. According to the transformation rule for 4 momentum, perpendicular momentum transferred remains unchanged while parallel momentum transferred scales by $\gamma$. So we have: $dp'=(\gamma Fdt\cos\alpha, Fdt\sin\alpha)$. Then:

$$ F' = \frac{dp'}{dt'} = \frac{\gamma Fdt\cos\alpha, Fdt\sin\alpha}{\gamma dt} = (F\cos\alpha, \frac{1}{\gamma}F\sin\alpha) $$

So the force exerted by the stick perpendicular to the direction of motion has been scaled by a factor of $1/\gamma$, just like the perpendicular electromagnetic force was, while parallel force remains the same. Mystery solved, but in a very strange way: The moving stick is exerting a force along a different axis than the one it's pointing in. But that's just special relativity for you. (Length contraction doesn't compensate for this, in fact it makes the stick point in a direction even farther from the direction it transfers force in!)

According to the Lorentz transformation formula for the electric field, we have:

$$ \mathbf E_\perp' = \gamma(\mathbf E_\perp+\mathbf v\times \mathbf B_\perp) $$

Since $\mathbf B=0$ in the rest frame, in the moving frame $\mathbf E$ becomes stronger by a factor of $\gamma$. We might hope for these effects to cancel out. The size of the magnetic field generated is $|B_\perp| = \gamma v|E_\perp|/c^2$, by a similar Lorentz transformation law.

Lorentz force on the moving charge will be $qv|B_\perp|$, since the field is perpendicular to the direction of motion. So the overall force due to magnetic field is $\gamma |E_\perp| qv^2/c^2$. Meanwhile, the magnitude of the electric force on the charge in the direction perpendicular to motion is $\gamma |E_\perp| q$. Since these are pointing in opposite directions, the net attractive force, perpendicular to the direction of motion, is:

$$ \gamma |E_\perp|q(1-\frac{v^2}{c^2}) = \frac{1}{\sqrt{1-v^2/c^2}}|E_\perp|q(1-v^2/c^2) = |E_\perp|q\sqrt{1-v^2/c^2} $$

But this is not a complete cancellation, the perpendicular electromagnetic force is still reduced in the moving frame. The only hope now is that maybe in the moving frame, the insulating stick applies a force that isn't parallel to itself.

In the resting frame, the stick transmits a force $F$ between the two charges. The definition of force is $dp/dt$: the rate of flow of momentum. In the resting frame, both charges have 0 momentum at all times. So the net momentum change is 0. Some momentum is flowing through the electric field between the charges at a rate of $F$, and an equal and opposite quantity of momentum is flowing through the insulating stick, also at a rate of $F$.

Let's consider the change in force applied by the stick after our Lorentz transformation. In the rest frame, we have time coordinate $t$. In the moving frame, we have time coordinate $t' = \gamma(t-vx/c^2)$. At rest, $x=0$ at all times, so $t' = \gamma t$. This means that a small interval of time $dt$ in the rest frame corresponds to a slightly longer interval of time in the moving frame, $dt'=\gamma dt$. The momentum transferred in a time $dt$ is $dp = (Fdt\cos\alpha, Fdt\sin\alpha)$ in the rest frame. According to the transformation rule for 4 momentum, perpendicular momentum transferred remains unchanged while parallel momentum transferred scales by $\gamma$. So we have: $dp'=(\gamma Fdt\cos\alpha, Fdt\sin\alpha)$. Then:

$$ F' = \frac{dp'}{dt'} = \frac{\gamma Fdt\cos\alpha, Fdt\sin\alpha}{\gamma dt} = (F\cos\alpha, \frac{1}{\gamma}F\sin\alpha) $$

So the force exerted by the stick perpendicular to the direction of motion has been scaled by a factor of $1/\gamma$, just like the perpendicular electromagnetic force was, while parallel force remains the same. Mystery solved, but in a very strange way: The moving stick is exerting a force along a different axis than the one it's pointing in. But that's just special relativity for you. (Length contraction doesn't compensate for this, in fact it makes the stick even less parallel to the force it transfers!)