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

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 $Q=\lambda Id$.