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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 $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth 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 $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; 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 $a_1{e}_3$. 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 parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts 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 $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.