Is $A^T A$ similar to $AA^T$?

If $A$ is a square real matrix and let $A=U D V^T$ be the SVD decomposition.

$$A^TA=VD^2V^T$$

$$AA^T=UD^2U^T$$

Notice that $$(UV^T)A^TA(VU^T)=AA^T$$

In general the statement is false. E.g. consider the complex matrix $A=\pmatrix{1&i\\ 0&0}$, for which $AA^T=0\ne A^TA$. However, the statement is true for real square matrices. A proof was given in the other answer here, but that proof can actually be made simpler if you are allowed to use polar decomposition: let $A=PU$, where $P$ is symmetric positive semidefinite and $U$ is real orthogonal (so that $U^T=U^{-1}$). Then $A^TA=U^TP^2U$ is similar to $AA^T=P^2$.