Restriction of an irreducible character to a normal subgroup Let $K$ be a finite group, $\Sigma$ a subgroup of $Aut(K)$. Let $\psi$ be an irreducible character of $K\rtimes \Sigma$. We want to show the following equality:

$\psi|_K = \sum_{\chi \in \mathcal{O}} \langle \psi|_K, \chi\rangle \cdot \chi$

Where the sum is over all irreducible characters of $K$ that lie in a unique orbit $\mathcal{O}$ under the action of $\Sigma$. According to Schur orthogonality, a prior this sum should be over all irreducible characters of $K$. But this is saying essentially that outside of a unique orbit of the $\Sigma$ action, no other characters appear in the restriction $\psi|_K$.
Does anyone have a proof of this fact?

I think the following works. I am assuming that everything is over $\Bbb{C}$. Let $G=K\rtimes\Sigma$.
Assume that $\psi$ is afforded by the $G$-module $V$. $V\vert_K$ is usually not irreducible as a $K$-module, but it has a non-trivial irreducible $K$-submodule $W$. Let $\eta$ be the character of $W$. 
Let $\sigma\in\Sigma$ be arbitrary. Consider the subspace $\sigma\cdot W$. Because $\sigma$ normalizes $K$, we see that $\sigma\cdot W$ is also a $K$-submodule of $V$: for all $k\in K, w\in W$, we have
$$
k\cdot(\sigma\cdot w)=\sigma(\sigma^{-1}k\sigma\cdot w)=\sigma\cdot(k^\sigma\cdot w)\in\sigma\cdot W.
$$
By the same calculation, the $K$-character afforded by $\sigma\cdot W$ is
$\eta^\sigma$ defined by $\eta^\sigma(k)=\eta(k^\sigma)$. That is, another character of $K$ in the $\Sigma$-orbit of $\eta$.
Consider the sum of $K$-modules 
$$
U=\sum_{\sigma\in\Sigma}\sigma\cdot W\subseteq V.
$$
Clearly $U$ is stable under the action of $\Sigma$ as well. Therefore it is stable under the action of all of $G$. 
But we assumed that $V$ is an irreducible $G$-module. Therefore $U=V$. The claim follows from this because the $K$-character afforded by $U$ is a sum of characters of the form $\eta^\sigma$.
