$\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\C}{\mathbf C}$
Let $G$ be a compact Hausdorff group and $\mu$ be the normalized Haar measure on $G$. Let $\sigma:G\to \mathscr U(V)$ be a finite dimensional unitary representation of $G$. We will write $\sigma_g$ to denote $\sigma(g)$.
Let $v, w\in V$ be fixed, and define a map $T:V\to V$ by writing $T(x) = \ab{x, v}w$. Consider the linear operator $$ \tilde T = \int_G \sigma_{g^{-1}} \circ T\circ \sigma_g\ d\mu(g) $$ Then $\tilde T$ is a $G$-equivariant endomorphism of $V$, and therefore by Schur's lemma there is $\lambda\in \C$ such that $\tilde T = \lambda I$. But then we have, for any $v', w'\in V$ that $$ \ab{\tilde Tv', w'} = \lambda \ab{v', w'} $$ giving $$ \int_G \ab{\sigma_g v', v}\overline{\ab{\sigma_g w', w}}\ d\mu(g) = \lambda \ab{v', w'} $$
Question. Is $\lambda$ same as $\ab{v, w}$?
The $G$-averaging process (Weyl's unitary trick) doesn't change the trace, since $\sigma_g$s are all unitary:
$$ \mathrm{tr}(\overline{T})=\int_G \mathrm{tr}(\sigma_g^{-1}T\sigma_g)=\int_G \mathrm{tr}(T)=\mathrm{tr}(T). $$
On the one hand, $\mathrm{tr}(\overline{T})=(\dim V)\lambda$. On the other hand, $\mathrm{tr}(T)=\mathrm{tr}(wv^\dagger)=\langle v,w\rangle$.
So, actually, $\lambda=\langle v,w\rangle/\dim V$.