= Solution
Write the jump from side $-$ to side $+$ as $[\mathbf F]=\mathbf F_+-\mathbf F_-$. A thin rectangular loop in the static <Faraday's law> and a thin pillbox in <Gauss's law for magnetism> give
$$
\boxed{\mathbf n\mathbin\times[\mathbf E]=0,
\qquad
\mathbf n\mathbin\cdot[\mathbf B]=0.}
$$
Thus the tangential electric field and normal magnetic field are continuous in the plane's rest frame.
For the boost $\mathbf v=v\mathbf n$, the <Lorentz transformation of electromagnetic fields> gives
$$
\mathbf E'_t=\gamma(\mathbf E_t+\mathbf v\mathbin\times\mathbf B),
\qquad
B'_n=B_n.
$$
Therefore
$$
\boxed{[B'_n]=0.}
$$
Moreover $[\mathbf E]$ is purely normal, so $[\mathbf B'_t]=\gamma[\mathbf B_t]$. Consequently the transformed tangential boundary condition is
$$
\boxed{[\mathbf E'_t]=\mathbf v\mathbin\times[\mathbf B'_t].}
$$
Using the <magnetic-field jump across a surface current>, $\mathbf n\times[\mathbf B]=\mu_0\mathbf K$, this can equivalently be written
$$
\boxed{[\mathbf E'_t]=\mu_0\gamma v\mathbf K.}
$$
The nonzero jump is consistent with Faraday's law because the magnetic-field discontinuity itself moves through Albert's frame.
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