= Solution
Fix the two subsystem dimensions. For <reflexivity>, $|\Psi\rangle=(I_d\otimes I_{d'})|\Psi\rangle$, and both <identity operators> are unitary.
For symmetry, if $|\Psi\rangle=(U\otimes V)|\Phi\rangle$, apply the inverse <local unitary transformation>:
$$
|\Phi\rangle=(U^\dagger\otimes V^\dagger)|\Psi\rangle.
$$
The <conjugate transposes> of <unitary operators> are unitary, so the reverse relation holds.
For <transitivity>, if additionally $|\Phi\rangle=(R\otimes S)|\Theta\rangle$, the <tensor product> multiplication rule gives
$$
|\Psi\rangle=(U\otimes V)(R\otimes S)|\Theta\rangle=(UR\otimes VS)|\Theta\rangle.
$$
Both products are unitary, for example $(UR)^\dagger(UR)=R^\dagger U^\dagger UR=I$. Thus \b[<local-unitary equivalence> is reflexive, symmetric and transitive], and hence is an <equivalence relation>.
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