Is the orthogonal polar factor the unique submersion satisfying an orthogonality relation? $\newcommand{\psym}{\text{Psym}_n}$
$\newcommand{\sym}{\text{sym}}$
$\newcommand{\Sym}{\operatorname{Sym}}$
$\newcommand{\Skew}{\operatorname{Skew}}$
$\renewcommand{\skew}{\operatorname{skew}}$
$\newcommand{\GLp}{\operatorname{GL}_n^+}$
$\newcommand{\SO}{\operatorname{SO}_n}$
The orthogonal polar factor map $O:\GLp \to \SO$, defined by requiring $A=
O(A)P$ for some symmetric positive-definite $P$, is a smooth submersion satisfying $A \perp T_{O(A)}\SO$.

Question: Let $F:\GLp \to \SO$ be a smooth submersion satisfying $A \perp T_{F(A)}\SO$. Does $F(A)=Q \cdot O(A)$ or $F(A)= O(A)  \cdot Q$ for some $Q \in \SO$? 

Edit: An equivalent reformulation of the question:
$A \perp T_{F(A)}\SO=F(A)\skew \iff A \in (F(A)\skew)^\perp=F(A)(\skew)^\perp=F(A)\sym$. Thus, if we define $S(A)=F(A)^{-1}A$, $S:\GLp \to \sym$ is smooth. 

So, a submersion $F:\GLp \to \SO$ satisfies the orthogonality requirement if and only if there exist a smooth map $S:\GLp \to \sym$, satisfying $A=F(A)S(A)$. 

Using the polar decomposition, we have $O(A)P(A)=A=F(A)S(A)$, so $S(A)=Q(A)P(A)$ where $Q(A)=F(A)^{-1}O(A)$. We want to prove that $Q:\GLp \to \SO$ is constant. I will now prove that for matrices $A$ having distinct singular values, $Q(A)$ can obtain a finite number of values. (The set of admissible values depends on $A$). I am quite sure this fact can be used to force $Q$ to be constant or at least something very constrained, but I am not sure how. Here is the proof:
Since $S(A)=Q(A)P(A) \in \sym$ , we have $PQ^T=(QP)^T=S^T=S=QP$. By orthogonally diagonalizing $P$, we can write $P=V\Sigma V^T$, so we now have
$$ V\Sigma V^T Q^T=QV\Sigma V^T \Rightarrow \Sigma V^T Q^TV=V^TQV\Sigma. $$ 
Setting $\tilde Q=V^TQV$ we thus have $ \Sigma \tilde Q^T= \tilde Q \Sigma$ where $\tilde Q \in \SO$. Since we assumed that the singular values of $A$ are distinct (i.e. the diagonal entries of $\Sigma$ are distinct), an explicit calculation now shows that $ \tilde Q$ must be diagonal. Since it is also orthogonal, we must have $\tilde Q_{ii}=\pm 1$ for all $i$. So, $\tilde Q$ can assume a finite set of values; (which implies the same thing for $ Q$).

Comment: I am not sure for which $Q \in \SO$ $F(A)=Q\cdot O(A)$ satisfies the requirement. A necessary condition is $Q^2=Id$; I don't know if it's sufficient.
Indeed, let $Q \in \SO$, and set $F(A)=Q\cdot O(A)$. Then $ A \perp T_{F(A)}\SO=T_{Q\cdot O(A)}\SO=QT_{O(A)}\SO,$ so for $A=Id$ we have  $ Id \perp Q\skew \Rightarrow Q^T \perp \skew \Rightarrow Q^T \in \sym \Rightarrow Q^2=Id$.
 A: I will use the notation from the question, $O(A)P(A)=A=F(A)S(A)$ and $S(A)=Q(A)P(A).$
The only possibilities are $Q(A)=I$ everywhere and $Q(A)=-I$ everywhere, with $-I$ only valid $n$ is even.
To show this I will first argue that $Q(I)=\pm I.$
The matrices $P(A),Q(A),S(A)$ all commute: diagonalize $S$ as $V\Sigma V^T$ with $V\in SO_n$ (possibly a different $V$ to the one in the question), then $P(A)=V|\Sigma| V^T$ and $Q(A)=V\operatorname{sgn}(\Sigma) V^T$ where $|\cdot|$ and $\operatorname{sgn}$ are applied entrywise to the diagonal elements. This can be seen from the expression $FS=(FV\operatorname{sgn}(\Sigma)V^T)(V|\Sigma|V^T)=OP$ and uniqueness of orthogonal polar decomposition.
For the case $n=2,$ we can use the above expression for $Q(A)$ to get $Q(A)^2=I.$ And $Q(A)\in SO_n.$ This forces $Q(A)=\pm I.$
Let $D=D(\epsilon)=\operatorname{diag}(1+\epsilon,1+2\epsilon,\dots,1+n\epsilon),$ and let $U$ be an arbitrary element of $SO_n.$ Since $Q(UDU^T)$ commutes with $P(UDU^T)=UDU^T,$ the conjugate $U^T Q(UDU^T) U$ commutes with $D.$ Any matrix $M$ that commutes with $D$ must be diagonal because $(DM-MD)_{ij}=M_{ij}(D_{ii}-D_{jj})=0$ forces $M_{ij}=0$ for $i\neq j.$ Taking $\epsilon\to 0$ and applying continuity of $D\mapsto U^T Q(UDU^T) U$ we get that $U^T Q(I) U$ is diagonal. This restricts $U^T Q(I) U$ to a discrete set $\operatorname{diag}(\pm1,\dots,\pm 1).$ Since $SO_n$ is connected, $U\mapsto U^T Q(I) U$ is constantly $Q(I).$ In other words $Q(I)$ commutes with every special orthogonal matrix, but for $n>2$ that forces $Q(I)=\pm I$ as claimed.
From the above expression for $Q(A)$ it is clear that $Q(A)$ has eigenvalues $\pm 1,$ which means the trace is a discrete invariant and must therefore be constant. If $Q(I)=I$ then the trace is $n$ and $Q(A)=I$ everywhere, and if $Q(I)=-I$ then the trace is $-n$ and $Q(A)=-I$ everywhere.
