Difference between $T(A) = A$ and $T^{-1}(A) = A$

Question

I am a bit confused about the title.

Let $T:X\rightarrow X$ be a map where $X \subset \mathbb{R}^n$. Let $A\subset X$.

I know that $$T^{-1}(A) = \{x\in X: T(x)\in A\}.$$

and if $A$ is $T$-invariant, we have $$T(A) = A.$$ In this case, $A$ is a bit like an attractor.

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My questions are:

1. Suppose $T^{-1}(A)=A$, can we say $T(A) = A$? or we can just say $T(A)\subset A$?
2. Suppose $T(A) = A$, can we say $T^{-1}(A) = A$?

I think 2. is not correct; because if $A$ admits a basin of attraction of $A$, say $V$, such that $A\subset V\subset X$, then it could be possible that $A\subset T^{-1}(A)$.

Answer 1

Neither equality is necessarily true.

Let $X=\mathbb{R}$, and let $T:X\to X$ be given by $T(x)=x^2$.

For the first one, if $A=X$, then $T^{-1}(A)=A$, but $T(A)=[0,\infty)$ which is a proper subset of $A$.

For the second one, if $A=[0,\infty)$, then $T(A)=A$, but $T^{-1}(A)=X$ which properly contains $A$.