Is every eigenvector of AA an eigenvector of A?

Hint: Consider $$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ Or $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$

A rotation by 90° as represented by a matrix e.g. $$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$ has no real eigenvalue ($\chi_A(\lambda)=\lambda^2+1$ is the characteristic polynomial) and so no eigenvectors. The rotation by 180° $$A^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$ has the whole space as a space of eigenvectors and zero vector .

No, it is false: consider in a real vector space $$A=\begin{bmatrix} 0&1\\1&0\end{bmatrix}\implies A\circ A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$ Then the vector $\begin{bmatrix} 1\\0\end{bmatrix}$ is an eigenvector of $A^2$ and not of $A$.