A simple equality related to an irreducible representation of a finite group Let $G$ be a finite group of cardinal $|G|$. Let $u \colon G \to M_n$ be an irreducible unitary representation of $G$. How to show that for any matrix $A \in M_n$ we have
$$
\frac{1}{|G|} \sum_{g \in G} u_g A u_g^*
=\frac{\mathrm{Tr(A)}}{n} \mathrm{I} \ ?
$$
It is maybe a consequence of orthogonality relations ?
 A: Let $B = \sum_{g \in G} u_g A u_g^{-1}$. This is an intertwiner for the action of $G$, since for any fixed $x \in G$ we have
$$ u_x B = \sum_{g \in G} u_{xg} A u_{g}^{-1} = \sum_{h \in G} u_hAu_{x^{-1} h}^{-1} = \sum_{h \in G} u_h A u_h^{-1} u_x = B u_x,$$
where we have re-indexed the sum using the bijection $h = xg$. Since $B$ is an intertwiner of the irreducible representation $u$, it must be equal to some scalar times the identity: we have $B = \lambda I$ for some $\lambda \in \mathbb{C}$. We can find that scalar by taking the trace:
$$ \operatorname{tr}(B) = \sum_{g \in G} \operatorname{tr}(u_g A u_g^{-1}) = \sum_{g \in G} \operatorname{tr}(u_g^{-1} u_g A) = \sum_{g \in G} \operatorname{tr}(A) = |G| \operatorname{tr}(A).$$
On the other hand since $B = \lambda I$ we have $\operatorname{tr}(B) = n \lambda$, and conclude that $\lambda= \frac{|G|}{n} \operatorname{tr}(A)$, making $B = \frac{|G|}{n} \operatorname{tr}(A) I$. In your case the identity follows because when $u$ is unitary, $u_g^{-1} = u_g^*$ for all $g \in G$.
