 First-order cancellation for an exchange-odd perturbation

Let $X$ be a unitary operator with $X^2=I$, and let $P$ be the orthogonal projection operator onto an unperturbed energy eigenspace. Suppose $X$ acts as the same sign $\epsilon =\pm 1$ throughout that eigenspace and $XVX=-V$. For any $u,v$ in the eigenspace, preservation of the inner product gives $⟨u,Vv⟩=⟨Xu,XVv⟩=-⟨\epsilon u,V\epsilon v⟩=-⟨u,Vv⟩$. Hence every projected matrix element vanishes, so $PVP=0$ and degenerate perturbation theory gives zero first-order energy corrections. A zero diagonal alone would be insufficient. For equal spins, the exchange parity of two coupled identical spins is constant on each total-spin sector, while their spin-projection difference changes sign under exchange.