Number of orbits of irreducible representations equals number of orbits of conjugacy classes under any subgroup of automorphism group

Statement
Suppose $$G$$ is a finite group, $$K$$ is a splitting field for $$G$$, and $$A$$ is a subgroup of the automorphism group of $$G$$. Denote by $$C(G)$$ the set of conjugacy classes in $$G$$ and by $$R(G)$$ the set of (equivalence classes of) irreducible representations of $$G$$ over $$K$$. Then, $$A$$ acts naturally on both $$C(G)$$ and $$R(G)$$.

The claim is that the number of orbits under the action of $$A$$ on $$R(G)$$ equals the number of orbits under the action of $$A$$ on $$C(G)$$.

Related facts

 * Number of irreducible representations equals number of conjugacy classes (extreme case where the subgroup has no outer automorphisms)
 * Number of orbits of irreducible representations equals number of orbits under automorphism group (extreme case where the subgroup is the whole automorphism group).
 * Cyclic quotient of automorphism group by class-preserving automorphism group implies same orbit sizes of conjugacy classes and irreducible representations under automorphism group

Facts used

 * 1) uses::Application of Brauer's permutation lemma to group automorphism on conjugacy classes and irreducible representations, which in turn uses uses::Brauer's permutation lemma (actually, we don't need the full strength of the statement about cycle types for our purpose, we simply need the fixed point version, which can be deduced even more directly)
 * 2) uses::Orbit-counting theorem (also called Burnside's lemma)

Proof
Given: A finite group $$G$$, a splitting field $$K$$ for $$G$$, a subgroup $$A$$ of the automorphism group of $$G$$. $$C(G)$$ denotes the set of conjugacy classes of $$G$$, $$R(G)$$ denotes the set of equivalence classes of irreducible representations of $$G$$ over $$K$$.

To prove: The number of orbits of $$C(G)$$ under the natural action of $$A$$ equals the number of orbits of $$R(G)$$ under the action of $$A$$.

Proof: