To what extent are the factors in a matrix equivalence unique? Suppose we have $n\times m$  matrix $Q$, and that $\exists$ $ n\times n$ matrices $P,P'$, and $m \times m$ matrices $R,R'$ all invertible, such that $PQR = \begin{bmatrix} I_{r\times r} & 0 \\ 0 & 0\end{bmatrix}= P'QR'$. Where $r$ is the rank of the matrix $Q$.
What is the relationship between $P,P',R,R'$? I know this question is not specific enough, but I'm not sure how to frame what I'm looking for properly. I want to know to what extent we can consider the factors in this decomposition unique.
 A: First of all, it is helpful to note that this is equivalent to considering a rank-factorization of $Q$. In particular, rewrite this in the form
$$
 Q  = A\pmatrix{I_r & 0\\0 & 0}B
$$
(in particular, we can take $A = P^{-1}$ and $B = Q^{-1}$).
if we partition $A$ into two block-columns and $B$ into two block-rows, then we have
$$
Q = A\pmatrix{I_r & 0\\0 &0}B \iff
Q = \pmatrix{A_1 & A_2}\pmatrix{I_r & 0\\0 &0} \pmatrix{B_1 \\ B_2}\\
\iff Q = A_1 I_r B_1 = A_1 B_1,
$$
where we note that $A_1$ has linearly independent column, and $B_1$ has linearly independent rows.  Once $A_1$ and $B_1$ are chosen, we need only complete the columns of $A_1$ and the rows of $B_1$ to form bases of $\Bbb R^m$ and $\Bbb R^n$.
Now, the question becomes to what extent are the choice of $F = A_1$ (size $m \times r$) and $G = B_1$ (size $r \times n$) unique in the factorization $Q = FG$?  We see that for any invertible size $r$ matrix $S$, taking $F' = FS$ and $G' = S^{-1}G$ gives us a new factorization. In fact, I claim that every rank factorization can be attained in this fashion. Indeed, this can be seen as a consequence of the uniqueness of the "skinny" QR factorization.
