# For which $P,Q \in \text{SO}$ $T_P\text{SO}$ and $T_Q\text{SO}$ are parallel?

I am curious: For which $$P,Q \in \text{SO}_n$$ does $$T_Q\text{SO}_n=T_P\text{SO}_n$$ hold?

This reduces to the question at the identity,i.e. for which $$Q \in \text{SO}_n$$, $$T_Q\text{SO}_n=T_{Id}\text{SO}_n=\text{skew}$$. I will now prove that $$Q^2=Id$$ is a necessary condition. Is it sufficient?

Since $$T_Q\text{SO}_n=QT_{Id}\text{SO}_n=Q\text{skew}$$, this happens if and only if $$Q\text{skew}=\text{skew}$$, i.e. $$-AQ^T=(QA)^T=-QA$$ for every $$A \in \text{skew}$$, or $$AQ^T=QA$$. Taking traces we get $$\langle Q,A\rangle=\text{tr}(Q^TA)=\text{tr}(AQ^T)=\text{tr}(QA)=\text{tr}(AQ)=-\text{tr}(A^TQ)= \langle A,Q\rangle,$$

so $$\langle Q,A\rangle=0$$ for every $$A \in \text{skew}$$, i.e. $$Q \in \text{skew}^{\perp}=\text{sym}$$, so $$Q^T=Q$$, or $$Q^2=Id$$.

Note that at even dimensions $$Q=-Id$$ is always a solution. For dimension $$n=2$$, this is indeed the only non-trivial solution (since $$\text{SO}_2$$ is the circle). In that case $$Q^2=Id$$, and $$Q=\pm Id$$ are equivalent.

I think this works: the equation $$AQ^T=QA$$, along with symmetry of $$Q$$, means that $$Q$$ commutes with every skew-symmetric matrix. We can write any matrix $$M$$ as a sum $$M = S + R$$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $$[Q,M]$$:
$$$$\begin{split} Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\\ &=QS-SQ\\ &=QS - (QS)^T \end{split}$$$$ Where we used commutativity of $$Q$$ with skew-symmetric matrices, and then symmetry of $$Q$$ and $$S$$. The result in the last line is clearly antisymmetric, so $$Q$$ commutes with that as well: $$$$\begin{split} 0 &= Q(QS-SQ) - (QS-SQ)Q\\ &=QQS-QSQ-QSQ+SQQ\\ &=2S-2QSQ \end{split}$$$$ So that $$S = QSQ$$, or $$QS=SQ$$, since $$Q^2=Id$$. But then $$Q$$ commutes with the symmetric part of the arbitrary matrix $$M$$ as well, so in fact $$Q$$ commutes with every matrix! This implies that $$Q$$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $$Q=\lambda I$$ with $$\lambda^2 = 1$$, so $$Q=\pm I$$.
Suppose that $$AQ^T=QA$$ for every $$A \in \text{skew}$$. Then $$Q=\lambda Id$$. (Here we do not assume $$Q$$ is orthogonal).
Proof: The argument in the question shows $$Q^T=Q$$, so $$AQ=QA$$, i.e. $$Q$$ commutes with every $$A \in \text{skew}$$. By exponentiating, we get that $$Q$$ commutes with $$e^A$$ for every $$A \in \text{skew}$$, i.e. $$Q$$ commutes with all rotations. So, if $$n>2$$, then $$Q=\lambda Id$$. If $$n=2$$, then $$Q$$ must be a scaled rotation; since it is also symmetric, this forces $$Q=\lambda Id$$.