Cyclic quotient of automorphism group by class-preserving automorphism group implies same orbit sizes of conjugacy classes and irreducible representations under automorphism group

Statement
Suppose $$G$$ is a finite group and $$K$$ is a fact about::splitting field for $$G$$. Suppose the quotient group of the automorphism group $$\operatorname{Aut}(G)$$ by the normal subgroup comprising the class-preserving automorphisms is a cyclic group.

Then, $$G$$ is a fact about::finite group having the same orbit sizes of conjugacy classes and irreducible representations under automorphism group. The action of $$\operatorname{Aut}(G)$$ on $$G$$ induces actions both on the set of conjugacy classes $$C(G)$$ and the set of (equivalence classes of) irreducible representations of $$G$$ over $$K$$. The claim is that the sizes of orbits of these two actions are the same.

Related facts

 * Number of irreducible representations equals number of conjugacy classes
 * Number of orbits of irreducible representations equals number of orbits under automorphism group: For any finite group $$G$$, the number of automorphism classes of elements of $$G$$ equals the number of equivalence classes of irreducible representations of $$G$$ under the $$\operatorname{Aut}(G)$$-action. Note that this is despite the fact that the orbit structure may be very different.

Facts used

 * 1) uses::Application of Brauer's permutation lemma to group automorphism on conjugacy classes and irreducible representations

Proof
The proof follows directly from fact (1), and the observation that the orbit sizes under the permutation action of a cyclic group are the same as the cycle type of any permutation that generates that cyclic group. More elaboration needed