Relation between the eigenvectors of $A$ and $A^TA$

Let $A=U\Lambda V^\top$, $U$, $\Lambda$ and $V$ are respectively left eigenvectors, eigenvalues and right eigenvectors.

$$A^\top A=U\Lambda V^\top V \Lambda U^\top=U\Lambda^2 U^\top$$

So what are the eigenvalues of $A^\top A$ ?

There is no simple relation between eigenvectors of $A$ and of $A^*A$ unless $A$ is normal. For example, if $$ A=\pmatrix{0&1\\0&0}, $$ $A$ has only one dimensional eigenvector subspace spanned by $[1,0]^T$, but $A^*A=0$ so any nonzero 2-vector is its eigenvector. Even if $A$ is diagonalizable, the eigenspaces of $A$ and $A^*A$ can be completely different.

If $A$ is normal, then $A=UDU^*$ for a unitary $U$ and diagonal $D$, so $A^*A=U|D|^2U^*$ is the spectral decomposition of $A^*A$.

However, there is a simple relation between left singular vectors of $A$ and the eigenvectors of $A^*A$ as indicated in another answer.