If $T$ is a homeomorphism and $\Omega$ is invariant under $T$, can we infer that $T(\Omega)=\Omega$?

Question

Let $d\in\mathbb N$, $T$ be a homeomorphism of $\mathbb R^d$ onto $\mathbb R^d$ and $\Omega\subseteq\mathbb R^d$ be invariant under $T$, i.e. $T(\Omega)\subseteq\Omega$.

Can we somehow infer (maybe by assuming that $T$ is even a $C^1$-diffeomorphism and/or $\Omega$ is closed/open) that $T(\Omega)=\Omega$?

Answer 1

Hint : Consider $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $$T(x)=\frac{x}{2}$$

and $\Omega = B(0,1)$ (the ball centered of $0$ and of radius $1$, closed or open as you prefer).