Is there any connection between $trace(AB^\top B A^\top)$ and $trace(A^\top BB^\top A)$ in general?

Question

Let $A$ and $B$ be recangular real matrices of the same shape.

Question 1. Is there any connection between $trace(AB^\top B A^\top)$ and $trace(A^\top BB^\top A)$ ?

The answer is probably, No! in general (an exception being when one of $A$ and $B$ is proportional to an orthogonal projection matrix), but maybe I'm missing something.

Question 2. What is a general condition between $A$ and $B$, for which the answer to Question 1 is affirmative ?

Answer 1

Following up on Ben's observation, given SVDs $A=U\Sigma V'$ and $B = R\Lambda S'$ we have

$$\begin{aligned} \|AB'\|_F^2 &= \|U\Sigma V'S\Lambda'R'\|_F^2 = \|\Sigma V'S\Lambda'\|_F^2 \\ \|B'A\|_F^2 &= \|S\Lambda'R'U\Sigma V'\|_F^2 = \|\Lambda'R'U\Sigma \|_F^2 \\ \end{aligned}$$

from this, if both $V'S$ and $R'U$ are diagonal, both norms are the same. This happens for instance when $S=V$ and $R=U$, i.e. precisely when $A$ and $B$ are simultaneously SVD-decomposable.

Recall that two square matrices are simultaneously diagonalizable if and only if they commute. It appears there is a similar condition for the existence of a simultaneous SVD decomposition:

Prove that if $A^TB$ and $AB^T$ are symmetric, then there exist orthogonal $U,V$ such that $UAV^T$ and $UBV^T$ are diagonal

However, the cited paper states that:

Either one (but in general, not both) of the diagonal matrices may be further restricted to have no negative elements

So it is not quite a simultaneous SVD, but close enough for our purposes here.