gr.group theory - Orbit structures of conjugacy class set and irreducible representation set under automorphism group

By Brauer's permutation lemma, the permutation characters are always equal, but the representations need not be isomorphic. For instance, the non-abelian group of order 27 and exponent 9 provides an example. One condition for an equivalence for subgroups of the automorphism group is given in Isaacs's Character Theory textbook as theorem 13.24 on page 230–231:

If S is a solvable subgroup of Aut(G), and gcd(|S|,|G|)=1, then the permutation representations of S on Irr(G) and Cl(G) are isomorphic.

This will rarely directly answer your question as Aut(G) and G usually have common prime divisors, but perhaps the ideas will be useful to you. In particular, it describes a strengthening of your #2 which implies #1.

Let me know if you would like GAP code to verify the order 27 example. The action on classes has orbits of sizes 1, 1, 1, 2, 6 and the action on the irreducibles has orbits of sizes 1, 2, 2, 3, 3.

GAP code to check permutation isomorphism:

OnCharactersByGroupAutomorphism := function( pnt, act )
  return Character( UnderlyingCharacterTable( pnt ),
  pnt{FusionConjugacyClasses(act^-1)} );
end;;
OnCBGA := OnCharactersByGroupAutomorphism;;

g := ExtraspecialGroup(27,9);;
a := AutomorphismGroup(g);;
gensIrr := List( GeneratorsOfGroup(a), f ->
  PermListList( Irr(g), List( Irr(g), chi -> OnCBGA( chi, f ) ) ) );
gensCcl := List( GeneratorsOfGroup(a), f ->
  PermList( FusionConjugacyClasses(f) ) );
# perm iso?
fail <> RepresentativeAction( SymmetricGroup( NrConjugacyClasses( g ) ),
  gensCcl, gensIrr, OnTuples );

Some of what you asked for might be more along the lines of asking if the permutation groups generated by gensIrr and gensCcl are conjugate, so I chose an example where even the images are not conjugate. The example given below of G=2×2×2 is the smallest if you only want strict permutation (non-)isomorphism.