Automorphism group has linearly equivalent actions on set of conjugacy classes and set of irreducible representations

Statement
Suppose $$G$$ is a finite group, $$C(G)$$ is the set of conjugacy classes of $$G$$ and $$R(G)$$ is the set of equivalence classes of irreducible representations of $$G$$ over $$\mathbb{C}$$. The automorphism group $$\operatorname{Aut}(G)$$ acts as permutations on the sets $$C(G)$$ and $$R(G)$$. The claim is that these permutation actions, when viewed as linear representations over $$\mathbb{Q}$$, are linearly equivalent.

Note that since linearly equivalent not implies permutation-equivalent (the failure of the analogue of Brauer's permutation lemma for non-cyclic groups) the representations need not be equivalent as permutation representations.

Related notions

 * Finite group having the same orbit sizes of conjugacy classes and irreducible representations under automorphism group
 * Finite group whose automorphism group has equivalent actions on the sets of conjugacy classes and irreducible representations