Let $d\in\mathbb N$, $T$ be a homeomorphism from $\mathbb R^d$ onto $\mathbb R^d$ and $\Omega\subseteq\mathbb R^d$ be closed awith $T(\Omega)\subseteq\Omega$.
Assuming that $\Omega=\partial M$ for some $d$-dimensional properly embedded, Jack Lee showed in this answer that $T(\Omega)=\Omega$.
However, it's not apparent to me at which point of his argumentation we actually need that $\Omega$ is of this particular form and not just any closed subset of $\mathbb R^d$ which is invariant under $T$.
In the first part, he is assuming that $\Omega$ is connected. The idea is to use the fact that a connected set has no proper clopen subset. Since $T$ is a homeomorphism and $\Omega$ is $\mathbb R^d$-closed, it is clear to me that $T(\Omega)$ is $\mathbb R^d$-closed and hence (since $T(\Omega)=T(\Omega)\cap\Omega$) $\Omega$-closed.
While it is also clear to me that $\left.T\right|_\Omega$ is a homeomorphism from $\Omega$ to $T(\Omega)$ and hence an open map, it is not clear to me how he's concluding that $T(\Omega)$ is $\Omega$-open. The fact that $\left.T\right|_\Omega$ is a homeomorphism from $\Omega$ to $T(\Omega)$ should only yield that $T(\Omega)$ is $T(\Omega)$-open, which is a triviality.
Take $T(x)=x/2$ on $\mathbb{R}$, take $\Omega=[-1, 1]$. Then $T\Omega \subsetneq \Omega$.
In the situation of that question the fact that $T$ is diffeomorphism onto its image shows that it's an open map from $\Omega$ to $\Omega$. (For manifolds of same dimension, the inverse function theorem tells you that diffeomorphic image of an open set is open: given $p$ and $T(p)$, some neighborhood of $p$ is mapped homeomorphically to some neighborhood of $T(p)$, hence the image under $T$ of an open set is open). Here there is no such property, $T$ is not an open map from $\Omega$ to itself.