= Returned polarization
Use the same fixed laboratory $x,y$ axes for the incident and returning <Jones vectors>, with $x$ along the fast axis. Remove a common <phase factor> so that the one-pass <Jones matrix> is $J=\operatorname{diag}(1,i)$. The reverse pass through this reciprocal plate accumulates the same relative <phase retardation>, not its inverse. The mirror has scalar reflection coefficient $r_m$ in these fixed axes. Therefore
$$
\boxed{\mathbf e_{\rm return}=J(r_mI)J\mathbf e_{\rm in}
=r_m\begin{pmatrix}1&0\\0&-1\end{pmatrix}\mathbf e_{\rm in}.}
$$
The effective retardance is $\pi$: the complete return acts as a <half-wave plate>. For initial <linear polarization> at angle $\alpha$ to the fast axis,
$$
\begin{pmatrix}\cos\alpha\\\sin\alpha\end{pmatrix}
\longmapsto
\begin{pmatrix}\cos\alpha\\-\sin\alpha\end{pmatrix},
\qquad \boxed{\alpha_{\rm return}=-\alpha\pmod\pi.}
$$
In particular, the $45^\circ$ input used for linear-to-circular conversion returns in <linear polarization> perpendicular to its original line: $(1,1)/\sqrt2\mapsto(1,-1)/\sqrt2$. An input along either plate axis returns along that same axis.
The wording does not specify the initial <polarization of waves>. Thus “orthogonal linear return” requires the usual $45^\circ$ linearly polarized input; it is not a universal answer for arbitrary polarized input. The <Jones matrix> above supplies the general answer. A non-normal metallic reflection may introduce different complex reflection coefficients for the two components and would require additional geometry and optical data. Descriptions of the intermediate <circular polarization> must also account for the reversed propagation direction when assigning handedness.
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