 Characteristic polynomials of AB and BA

For square matrices $A,B$ of the same size over any field,If $A$ is invertible, $BA=A^{-1}(AB)A$, so the products are similar matrices. To cover singular $A$, introduce an indeterminate $s$ and work over the rational function field $F(s)$. The matrix $A-sI$ is invertible there, hence $(A-sI)B$ and $B(A-sI)$ have the same characteristic polynomial. Both sides are polynomials in $s,t$, so specialize $s=0$. This proof also works over a finite field; choosing infinitely many scalars in the original field would not.

Equal characteristic polynomials do not force the products to be similar. With $A=\mathrm{diag}(1,0)$ and $B=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $AB=B\ne 0$ but $BA=0$. Both have characteristic polynomial $t^2$, while their minimal polynomials are $t^2$ and $t$.