Solution

1. Rest frame S'

Plates parallel to the x'y' plane, separated in z' by a distance h.
Uniform electric field:
$\displaystyle \mathbf{E}' = (0,\,0,\,E_0), with E_0 = \dfrac{\sigma}{\varepsilon_0} (SI)$.
Uniform leakage current: $\mathbf{j}' = (0,\,0,\,j)$.
There is no magnetic field $(\mathbf{B}' = 0)$by symmetry.

The capacitor moves with velocity $\mathbf{v} = v\,\hat{\mathbf{z}} (v = \beta c)$, parallel to the electric field and the leakage current.
$\gamma = 1/\sqrt{1-\beta^2}$

Electromagnetic fields (boost in z):

\begin{aligned} E_z &= E'_z = E_0, \quad \mathbf{E}_\perp = \gamma(\mathbf{E}'_\perp + \mathbf{v}\times\mathbf{B}') = 0,\ B_z &= B'_z = 0, \quad \mathbf{B}_\perp = \gamma\!\left(\mathbf{B}'_\perp - \frac{\mathbf{v}}{c^2}\times\mathbf{E}'\right) = 0. \end{aligned}

$\boxed{\mathbf{E} =(0,\,0,\,E_0),\qquad \mathbf{B} = 0}$

The electric field does not change relative to the rest system.

The four-current $(c\rho, \mathbf{j})$
transforms as a four-vector. In S':
$\rho' = 0, \mathbf{j}' = (0,0,j)$

\begin{aligned} \rho &= \gamma\!\left(\rho' + \frac{v}{c^2}j'_z\right) = \gamma\frac{v}{c^2}j,\ j_z &= \gamma\,(j'_z + v\rho') = \gamma\,j,\ j_x &= j'_x = 0,\quad j_y = 0. \end{aligned}

The volumetric current density increases:
$j_z = \gamma\,j$.

A net charge density appears in the dielectric:
$\rho = \gamma v j / c^2$

However, the total current I through a cross-section remains invariant because the area of the plates (perpendicular to the motion) contracts:
$A = A'/\gamma$.

$I = \int j_z\,dA = (\gamma j)\left(\frac{A'}{\gamma}\right) = j A' = I'$