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That's not necessarily correct. Consider the potential $$U(x)=A\cos(x)$$ and boundary conditions $$\psi(+a)=\psi(-a),$$ $$\psi'(+a)=\psi'(-a)$$ for some $a\in\mathbb R$. The potential is symmetric, so there exist simultaneous eigenstates of both the Hamiltonian $\hat T+\hat U$ and the parity operator. But most of the eigenstates of this Hamiltonian are ...