How do I find $\operatorname{det} T_Q$? 
Let $S$ be the space of all $n \times n$ real skew symmetric matrices and let $Q$ be a real orthogonal matrix. Consider the map $T_Q: S \to S$ defined by $$T_Q(X) = QXQ^T.$$ Find $\operatorname{det} T_Q$.

I thought about diagonalizing $Q$, but I don't think we know it is real diagonalizable. I can show it is an isometry using the Hilbert-Schmidt inner product, but I can't really relate it to the determinant of $Q$ (I've seen posts here that say the determinant should be $\operatorname{det}Q^{n-1}$). So all I know is that $\operatorname{det} T_Q = \pm1$. How would I find $\det T_Q$?

Partial Answer: I will focus on the case that $n$ is even. The odd case can be handled similarly.
As a consequence of the block-diagonalizability of skew-symmetric matrices and the fact that $Q = \exp(P)$ holds for some skew-symmetric matrix $P$, we can show that there exists an orthogonal matrix $W$ such that
$$
WQW^T = D:= \pmatrix{A_1\\ & \ddots \\ && A_k}, \quad A_k = \pmatrix{a_k & -b_k\\ b_k & a_k}.
$$
Note that $T_Q = T_W \circ T_D \circ T_W^{-1}$, so that $\det(T_Q) = \det(T_D)$. Decompose $D$ into a product $D = D_1 \cdots D_k$ where
$$
D_1 = \pmatrix{A_1 \\ & I \\ & & \ddots \\ &&& I}, \dots, \quad D_k = \pmatrix{I \\ & \ddots \\ & & I \\ &&& A_k}.
$$
Note that $T_D = T_{D_1} \circ \cdots \circ T_{D_k}$. With that, it suffices to determine $\det T_{D_k}$.

Another approach is to use the fact that there exists a skew-symmetric matrix $M$ for which $Q = \exp(M)$ and that $T_Q = \exp(C_M)$, where
$$
C_M(X) = MX - XM.
$$
From there, we can use the fact that
$$
\det(T_Q) = \det(\exp(C_M)) = \exp(\operatorname{tr}(C_M)).
$$
Once we show that $\operatorname{tr}(C_M) = (n-1)\operatorname{tr}(M)$, it follows that
$$
\begin{align}
\det(T_Q) &= \exp(\operatorname{tr}(C_M)) = \exp((n-1)\operatorname{tr}(M)) = \exp(\operatorname{tr}(M))^{n-1} 
\\ & = \det(\exp(M))^{n-1} = \det(Q)^{n-1}.
\end{align}
$$
