Let $UP, U'P'$ be $n\times n$ real matrices where $U,U'$ are orthogonal and $P,P'$ are positive definite (hence $UP,U'P'$ are invertible.)
Then $UPU'P' = U''P''$ by polar decomposition. My question is this:
Is $U'' = U'U''$ and $P'' = P'P''$?
If $n=1$, it trivially holds. However, does it hold for higher $n$?
Edit: I think my question can be rewritten as
Is polar decomposition map gives group isomorphism between $GL(n,\mathbb{R})$ and $O(n) \times \{$Positive Definite matrix $\}$?
$$U=P'=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ $$P=\begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix}$$ $$U'=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$$ so $$UPU'P'=\begin{pmatrix} -1 & 1 \\ 1 & 1 \\ \end{pmatrix}$$ but if $U''=UU'$ and $P''=PP'$ then $$U''P''=\begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix}$$
As noted in the comments, a random example is very likely to work. Alternatively your property holds if and only if $PU'=U'P$. My choice of $U'$ swaps rows of $P$ if multiplied on the left, columns if multiplied on the right, so I just picked any $P$ for which these two operations are different.