= Solution
Let $s=k^\mu x_\mu$. The <Lorenz gauge condition> on the <plane electromagnetic wave> gives
$$
k_\mu\epsilon^\mu=0.
$$
The vacuum <wave equation> gives $k_\mu k^\mu=0$ for a nonzero amplitude; equivalently $k^0=\omega/c$ and $|\mathbf k|=\omega/c$. Use the <gauge transformation> $A'^\mu=A^\mu+\partial^\mu\chi$ with
$$
\chi=-\lambda\cos s,\qquad \epsilon'^\mu=\epsilon^\mu+\lambda k^\mu.
$$
Because $\Box\chi=0$, this is a <residual Lorenz gauge symmetry> and $k_\mu\epsilon'^\mu=0$. For a propagating <plane electromagnetic wave>, $k^0\ne0$, so take $\lambda=-\epsilon^0/k^0$. Then
$$
\boxed{\epsilon'^0=0,\qquad \mathbf k\cdot\boldsymbol\epsilon'=0.}
$$
The second equality follows from $-k^0\epsilon'^0+\mathbf k\cdot\boldsymbol\epsilon'=0$. The <electric field> and <magnetic field> are unchanged by this <gauge transformation>. Thus <Lorenz gauge> permits a representative in <radiation gauge>; <Lorenz gauge> alone does not require the spatial amplitude of every representative to be transverse.
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