If $\operatorname{rank}(A)$ = $\operatorname{rank}(A^2)$, show that nullspace of $A$ = nullspace of $A^2$

Let $A$ be a square matrix.

If $\operatorname{rank}(A)$ = $\operatorname{rank}(A^2)$

Prove that nullspace of $A$ = nullspace of $A^2$

The first thing I notice is that this $\implies$ $\operatorname{nullity}(A)=\operatorname{nullity}(A^2)$

Then I am kinda stuck, any hints?

You should show that the nullspace of $A$ is contained in the nullspace of $A^2$. By your observation, the dimension of the two nullspaces must necessarily be the same, so $\text{nullspace} A \subseteq \text{nullspace}A^2$ necessarily gives that they are equal.

For any matrix $C$, let $\Psi(C)$ and $\mathscr{N} (C) $ denote the rank and null space of $C$ respectively.

Let $A$ be an $n\times n$ square matrix such that $\Psi(A) =\Psi(A^2)$. If $x\in\mathscr{N} (A) $, then one can show that $x\in\mathscr{N} (A^2)$. We thus have $\mathscr{N} (A)\subseteq\mathscr{N}(A^2)$. But since it is given that $\Psi(A) =\Psi(A^2)$, we have $\dim\mathscr{N} (A) =\dim\mathscr{N} (A^2)$. Therefore, we can conclude that $\mathscr{N} (A^2)\subseteq \mathscr{N} (A)$, and hence $\mathscr{N} (A)=\mathscr{N} (A^2)$.