Consider three nonzero matrices $A, B, C$ such that $ABB^t=CBB^t$. Then which of the following are true?
(A) $r(A) =r(C)$
(B) nonzero eigenvalues of $A$ and $C$ are identical
(C) $AB=CB$
(D) none of the above
Try:
Since $ABB^t=CBB^t\implies \det(ABB^t) =\det(CBB^t) $
$\implies \det(A) \det(B) \det(B^t) =\det(C) \det(B) \det(B^t)$
$\implies \det(A) (\det(B) ^2=\det(C)(\det(B))^2$ [since $\det(B) =\det(B^t)$].
The above implies that $\det(A) =\det(C)$ which is possible iff they have the same set of nonzero eigenvalues and all the eigenvalues are nonzero . But if they have a 0 eigenvalue then we can't claim that. How to conclude?
Items (a) and (b) are certainly false. Take $A=B=\begin{pmatrix}1&0\\ \:0&0\end{pmatrix},C=\begin{pmatrix}1&0\\ \:0&2\end{pmatrix}$. Item (c) is true, and it relies on the fact that $B$ and $BB^t$ have the same column space. To see this, assume $B$ is $m\times n$, and fix $x\in \mathbb{R}^n$. Since $\text{Col}(B)=\text{Col}(BB^t)$ we can find $y\in \mathbb{R}^m$ such that $Bx=BB^ty$. So we have $$ABx=ABB^ty=CBB^ty=CBx$$ Since $x\in \mathbb{R}^n$ was chosen arbitrarily we must have $AB=CB$.