Determinant of the linear map given by conjugation. Let $S$ denote the space of skew-symmetric $n\times n$ real matrices, where every element $A\in S$ satisfies $A^T+A = 0$. 
Let $M$ denote an orthogonal $n\times n$ matrix, and $L_M$ denotes the linear map
$$L_M: A\mapsto MAM^{-1}.$$
What is the determinant of $L_M$ as an endomorphism of $S$?
 A: If we stack the entries of the matrices row by row, then $L_M=M\bigotimes M^{-T}:M_n\rightarrow M_n$ is a Kronecker product. The eigenvalues of $L_M$ are the $(\lambda_j/\lambda_k)_{j,k}$ where $\operatorname{spectrum}(M)=(\lambda_j=\exp(i\theta_j))_j$. Finally $\operatorname{spectrum}(L_M)=(\exp(i(\theta_j-\theta_k))_{j,k}$ and each eigenvalue has modulus $1$.
Here we consider $\phi={L_M}_{|S}$; then $\det(\phi)$ is a product (the result is real) of a part of the eigenvalues of $L_M$. Thus $\det(\phi)=\pm 1$. For example, let $n=2$ ; if $M\in O^+(2)$ then $\det(\phi)=1$ ($\phi(s)=s$) and if $M\in O^-(2)$, then $\det(\phi)=-1$ ($\phi(s)=-s$).
EDIT: on can slightly improve the previous result: ${L_M}^T=M^T\bigotimes M^T$ implies that $L_M{L_M}^T=MM^T\bigotimes MM^T=I_{n^2}$ ; moreover $\det(L_M)=\det(M)^{2n}=1$. Then $L_M\in O^+(n^2)$ and $\phi\in O^+(S)$ or $O^-(S)$.
A: Proposition For any orthogonal $n \times n$ matrix $M$, the conjugation map $L_M: S \to S$, $M \mapsto M A M^{-1}$ has determinant $1$ if $M \in SO(n)$ and/or $n$ is odd, and $-1$ if $M \in O(n) - SO(n)$ and $n$ is even.
Lemma For any orthogonal $n \times n$ matrix $M$, $\Phi: M \mapsto \det L_M$ is a homomorphism $O(n) \to \mathbb{R}^*$, that is, $\Phi$ satisfies:


*

*$\Phi(I_n) = 1$,

*$\Phi(MN) = \Phi(M) \Phi(N)$ for all $M, N \in O(n)$, and

*$\Phi(M^{-1}) = \Phi(M)^{-1}$ for all $M \in O(n)$.


Proof of Lemma This is immediate, because $\Phi$ is a composition of the determinant representation $GL(S) \to \mathbb{R}^*$ and the adjoint representation $O(n) \to GL(S)$, but for readers who aren't familiar with these ideas, we can verify these claims directly.


*

*We have $L_{I_n}(A) = I_n A I_n^{-1} = A$, so $L_{I_n}$ is the identity map on $S$, and thus $\Phi(I_n) = \det L_{I_n} = 1$.

*By definition, $L_{MN}(A) = (MN)A(MN)^{-1} = M(NAN^{-1})M^{-1} = L_M(L_N(A))$, so $\Phi(MN) = \det L_{MN} = \det (L_M \circ L_n) = \det L_M \det L_N = \Phi(M) \Phi(N)$.

*(Follows from (1) and (2).) $\square$


Proof of Proposition Since $\Phi$ is continuous and $O(n)$ is compact, the image of $\Phi$ is compact, and hence it is contained in $\{\pm 1\}$. (Suppose not; then there is an element $A$ with $\Phi(A) \not\in \{\pm 1\}$, and by inverting [and using (3)] if necessary we can assume $|\Phi(A)| > 1$. Then, by (2) the image under $\Phi$ of the sequence $(A, A^2, A^3, \ldots)$ is $(\det A, (\det A)^2, (\det A)^3, \ldots)$, which is unbounded, a contradiction with the compactness of the image of $\Phi$.) Since $\Phi$ is continuous with discrete components, it is constant on connected components, and by (1) in the Lemma, $\Phi(M) = \Phi(I_n) = \det L_{I_n} = 1$ for $M$ in the connected component of $I_n$ in $O(n)$, namely $SO(n)$. (Notice how quickly we arrive at this result just using a few easy observations.)
Since $O(n) - SO(n)$ is connected and $\Phi$ is constant on connected components, to compute $\det L_M$ for $M$ in this set, it suffices to compute it for a single element (for each $n$). A simple choice is
$Z := \left(\begin{array}{cc} -1 & 0 \\ 0 & I_n \end{array}\right)$.
We compute $\det L_Z$ directly by computing the matrix representation of $L_Z$ in the basis $(F_{ij})_{i < j}$ of $S$, where $F_{ij}$ is the matrix with $(i, j)$ entry $1$, $(j, i)$ entry $-1$, and all other entries zero, so that
$(F_{ij})_{kl} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}$.
Then, $L_Z(F_{ij}) = Z F_{ij} Z^{-1}$ has entries
$(L_Z(F_{ij}))_{ab}
= \sum_{k, l} Z_{ak} (F_{ij})_{kl} Z_{lb}
= Z_{ai} Z_{jb} - Z_{aj} Z_{ib}$,
and by construction this is just the $((a, b), (i, j))$ entry $(L_Z)_{(a, b), (i, j)}$ of the matrix representation of $L_Z$ in the basis $(F_{ij})_{i < j}$. We show that this matrix is in fact diagonal, using that $Z_{11} = -1$, $Z_{aa} = 1$ for $a > 1$, and $Z_{ab} = 0$ for $a \neq b$:
When $a = i = 1$ (so that $b, j > 1$), we have
$(L_Z)_{(1, b), (1, j)} = Z_{11} Z_{jb} - Z_{1j} Z_{1b} = -\delta_{jb}$;
when $i = 1$ and $a > 1$ (or, by symmetry of notation, $a = 1$ and $i > 1$, we have
$(L_Z)_{(a, b), (1, j)} = Z_{a1} Z_{jb} - Z_{aj} Z_{ib} = 0$;
when $a, i > 1$, we have
$(L_Z)_{(a, b), (i, j)} = Z_{ai} Z_{jb} - Z_{aj} Z_{ib} = \delta_{ai} \delta_{jb} - \delta_{aj} \delta_{ib} = \delta_{ai} \delta_{jb}$,
where the last equality comes from the fact that if $\delta_{aj} \neq 0$, then $a = j$, so that $i < j = a < b$, so $\delta_{ib} = 0$, that is the latter term in the second-to-last expression in the previous display is always zero.
In summary, the above computations give us that in the basis $(F_{ij})_{i < j}$, the matrix representation of $L_Z$ is diagonal, the $n - 1$ diagonal entries for the basis elements $F_{1j}$ are $-1$, and the remaining diagonal entries are all $1$. Thus, $\det L_Z = (-1)^{n - 1}$ as claimed. $\square$
