= Solution
On each half-space the <deformation> is affine. Continuity on the separating <plane> makes the two affine constants equal and gives $(A-B)t=0$ for every <tangent vector> $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ gives
$$
(A-B)x=[(A-B)N](N\cdot x),\qquad\boxed{A-B=a\otimes N,\quad a=(A-B)N.}
$$
This is the <Hadamard jump condition>. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the <matrix determinant lemma> gives
$$
\det B=\det A(1-N\cdot A^{-1}a).
$$
Both <determinants> are positive, hence $\boxed{A^{-1}a\cdot N<1}$.
Write $a_1=\eta_1$, $a_2=\eta_2$, $S=a_1^2+a_2^2$ and $\Delta=a_2^2-a_1^2\ne0$. The two proposed angles have $c=2a_1a_2/S$ and $s=\kappa\Delta/S$, and $c^2+s^2=1$. Therefore the displayed <rotations> lie in $SO(3)$. Direct multiplication gives
$$
RU_2-U_1=\frac\Delta S\begin{pmatrix}-a_2&-\kappa a_2&0\\\kappa a_1&a_1&0\\0&0&0\end{pmatrix}
=\boxed{\frac\Delta S(-a_2,\kappa a_1,0)\otimes(1,\kappa,0).}
$$
This is a <rank-one connection> for each sign. With a unit normal one may choose $N=(1,\kappa,0)/\sqrt2$ and jump vector $a=\sqrt2\Delta(-a_2,\kappa a_1,0)/S$. The two <rotations> are distinct because $\Delta\ne0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.
It remains to test the material <plane>. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the <plane> is spanned by $t$ and $e_3$. Both <deformation gradients> send $e_3$ to $a_1e_3$. If the <plane> is not the <twin plane>, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-<plane> vectors $Bt$ and $At$ are <linear dependence>[parallel]. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa)$ in their two-dimensional sections. Then
$$
\begin{aligned}
\det_2(Bt,At)
&=\frac\Delta S\det_2(Bt,d)(n\cdot t)\\
&=\boxed{\frac{\kappa a_1a_2\Delta}{S}(p_1-\kappa p_2)^2}.
\end{aligned}
$$
Its prefactor is nonzero, so <coplanarity> requires $p_1=\kappa p_2$, precisely $p\parallel(1,\kappa,0)\parallel N$. Conversely, on that <twin plane> $(A-B)x=0$, so the two <affine maps> agree and its image is a <plane>. Thus $\boxed{\Pi\text{ remains plane}\iff p\parallel N}$. All other allowed <planes> acquire a kink along their intersection with the twin. This is the <planarity test across a cubic-to-tetragonal twin>.
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