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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.

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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{equation} \begin{split} Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\\ &=QS-SQ\\ &=QS - (QS)^T \end{split} \end{equation} 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{equation} \begin{split} 0 &= Q(QS-SQ) - (QS-SQ)Q\\ &=QQS-QSQ-QSQ+SQQ\\ &=2S-2QSQ \end{split} \end{equation} 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 $Q=\pm I$.

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  • $\begingroup$ Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim. $\endgroup$ – Asaf Shachar Jan 7 at 12:58
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Here is a different approach to proceed: We shall in fact prove the following more general claim:

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 $Q=\lambda Id$.

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