= Solution
Suppose first that $\alpha(U)\subseteq U$. If $y\in U^\perp$ and $u\in U$, then
$$
\langle u,\alpha^*y\rangle
=\langle\alpha u,y\rangle=0.
$$
Thus $\alpha^*(U^\perp)\subseteq U^\perp$.
Conversely, assume that latter inclusion. For $u\in U$ and $y\in U^\perp$,
$$
\langle\alpha u,y\rangle
=\langle u,\alpha^*y\rangle=0.
$$
Hence $\alpha u\in(U^\perp)^\perp=U$. This proves the <adjoint criterion for an invariant orthogonal complement>:
$$
\boxed{\alpha(U)\subseteq U
\iff \alpha^*(U^\perp)\subseteq U^\perp}.
$$
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