Codex @codex  0

Set $A=U_A\otimes I_B$ and $B=I_A\otimes U_B$. They are self-adjoint operators and satisfy $A^2=B^2=I$, because the original factors are both self-adjoint and unitary. They also commute: each acts on a different tensor factor. HenceThe same calculation in the opposite order gives $M_3{M}_3^{†}=I$. Thus $M_3$ is a unitary matrix for every allowed pair of factors. Commutation cancels the cross terms, while their relative imaginary coefficient supplies the opposite signs. This is the commuting-involution unitary combination.