Is it true that $A^TA=A \implies A^2=A$? Is the converse true?

Note that $A^TA$ is a symmetric matrix, so if $A^TA=A$, then $A$ is symmetric. Consequently, $A^T=A$ which implies that $A^2=A^TA=A$. Conclusion: (1) is true.

(2) is false. Here's a counterexample $$A= \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} $$ Then $A^2=A$, but $A^TA\neq A$ (otherwise $A$ would be symmetric).

(1) is true:

If $A^{t}A=A$, then $A^{t}=(A^{t}A)^{t}=A^{t}A^{tt}=A^{t}A=A$. (Here, we have used the properties that $A^{tt}=A$ and $(AB)^{t}=B^{t}A^{t}$). It follows that $A^{2}=AA=A^{t}A=A$.

(2) is false:

Let $A=\begin{pmatrix}1 & 0\\ a & 0 \end{pmatrix}$, where $a$ can be any non-zero number. Then $A^{2}=A$. Now $A^{t}=\begin{pmatrix}1 & a\\ 0 & 0 \end{pmatrix}$, so $A^{t}A=\begin{pmatrix}1+a^{2} & 0\\ 0 & 0 \end{pmatrix}\neq A$.