Let $G$ be a group of "diagonal sign matrices $(g)$", i.e. a diagonal matrix with diagonal entries in $\{-1,1\}$. So $G$ is of cardinality $2^n$. For example: $$g = \begin{bmatrix}1 & 0 &0 \\0 & -1 & 0 \\0 & 0 & 1 \end{bmatrix}$$
Define $$P(X) = \frac{1}{2^n}\sum_{g\in G} gXg^T$$ with $X\in \mathbf{R}^{n\times n}$
My question: How to show $P$ is self-adjont?
So, obviously we have to show $P^T=P$:
$$P^T = \big(\frac{1}{2^n}\sum_{g\in G} gXg^T\big)^T = \frac{1}{2^n}\sum_{g\in G} \big(gXg^T\big)^T=\frac{1}{2^n}\sum_{g\in G} gX^Tg^T$$
How to go a step further?
http://arxiv.org/abs/1403.4914 (p.1326)
Choose some $h\in G$ and fix $X$. Then \begin{equation} hP(X)h^T = \frac{1}{2^n} \sum_{g\in G} hgXg^Th^T = \frac{1}{2^n} \sum_{g\in G} (hg)X(hg)^T = P(X). \end{equation} So we have \begin{equation} P(P(X)) = \frac{1}{2^n}\sum_{g\in G} gP(X)g^T = \frac{1}{2^n}\sum_{g\in G} P(X) = P(X), \end{equation} meaning that $P$ is idempotent. Some similar trickery will show that $PP^T = P^T$ and that $P^T P=P^T$. The fact that $PP^T = P^TP$ means that $P$ is a normal operator, and you can find arguments (e.g., here or here) showing that all normal, idempotent operators are self-adjoint.