= Solution
Define the reflected function $\phi(x)=\psi(-x)$. The <chain rule> gives $\phi'(x)=-\psi'(-x)$ and $\phi''(x)=\psi''(-x)$. Evaluating the <time-independent Schrödinger equation> for $\psi$ at $-x$ and using the <even function> identity $V(-x)=V(x)$ gives
$$
-\frac{\hbar^2}{2m}\phi''(x)+V(x)\phi(x)=E\phi(x).
$$
Thus reflection preserves the <energy eigenvalue>. The same substitution $y=-x$ in the <norm> gives
$$
\boxed{\int_{-\infty}^{\infty}|\phi(x)|^2\,dx=\int_{-\infty}^{\infty}|\psi(-x)|^2\,dx=\int_{-\infty}^{\infty}|\psi(y)|^2\,dy=1.}
$$
This is the action of the <parity operator> $R\psi(x)=\psi(-x)$, which is a <unitary operator>. For a reflection-symmetric physical problem it preserves the <Hamiltonian operator>'s <operator domain> as well as its differential expression.
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