# Character values up to permutation of characters need not determine automorphism class

## Statement

It is possible to have a finite group $G$ and a splitting field $K$ for $G$ in characteristic zero such that the following holds: There are elements $g_1,g_2 \in G$ in different conjugacy classes and a permutation $\sigma$ on the set of characters of irreducible representations (up to equivalence) of $G$ over $K$ such that the following holds:

• For every irreducible character $\chi$, we have $\chi(g_1) = \sigma(\chi)(g_2)$.
• $g_1$ and $g_2$ are not in the same automorphism class, i.e., they are not in the same orbit under the action of the automorphism group $\operatorname{Aut}(G)$ on $G$.

## Proof

### Example of the dihedral group

Further information: dihedral group:D8, linear representation theory of dihedral group:D8

This character table works over characteristic zero:

Representation/Conj class $\{e \}$ (size 1) $\{ a^2 \}$ (size 1) $\{ a, a^{-1} \}$ (size 2) $\{ x, a^2x \}$ (size 2) $\{ ax, a^3x \}$ (size 2)
Trivial representation 1 1 1 1 1
$\langle a \rangle$-kernel 1 1 1 -1 -1
$\langle a^2, x \rangle$-kernel 1 1 -1 1 -1
$\langle a^2, ax\rangle$-kernel 1 1 -1 -1 1
2-dimensional 2 -2 0 0 0

The same character table works over any characteristic not equal to 2 where the elements 1,-1,0,2,-2 are interpreted over the field.

Note that the element $a$ of order four and the element $x$ of order two have the same character values up to permutation of characters, but they cannot be in the same orbit under the action of the automorphism group because the elements have different orders.