Are the matrix products $AB$ and $BA$ similar?

$AB$ is conjugate to $BA$ if either $A$ or $B$ are invertible. If neither is the case, there are counterexamples: for example, it may be the case that $AB = 0$ while $BA \neq 0$. Explicitly, take $$A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right].$$

We have $AB = A$ but $BA = 0$.

However, there is a salvage: $AB$ and $BA$ have the same characteristic polynomial. See this blog post. (Short proof: this must hold if either $A$ or $B$ is invertible, and that condition is Zariski dense.)

If $A$ is invertible, then $AB = A(BA)A^{-1}$ which shows that $AB$ and $BA$ are similar. Similar (no pun intended) proof if $B$ is invertible.

As mentioned already, if either of $A$ or $B$ is invertible (and both are the same size), we have $$ \begin{align} AB=A(BA)A^{-1}\quad&\mbox{if $A$ is invertible}\\ AB=B^{-1}(BA)B\quad&\mbox{if $B$ is invertible} \end{align} $$

However, here is a short proof that even if $A$ is $m\times n$ and $B$ is $n\times m$, the characteristic polynomials of $AB$ and $BA$ differ only by a factor of $\lambda^{\large|n-m|}$.