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$\DeclareMathOperator\Out{Out}\DeclareMathOperator\CTS{CTS}$The outer automorphism group $\Out(G)$ of a finite group $G$ act on the character table by permuting columns (conjugacy classes) and rows (representations). As a result, there is an homomorphism $\psi:\Out(G)\rightarrow \CTS(G)$, where $\textrm{CTS}(G)$ is the symmetry group of the character table of $G$ (see Proposition 3.8.12 of Schedler - Group representation theory, lecture notes for reference). When is this homomorphism $\psi$ surjective or injective?

More specifically, given $\sigma\in\CTS(G)$, are there conditions involving properties of the conjugacy classes and representations that are being permuted by $\sigma$, that allows one to know that there exists $\alpha\in\Out(G)$ such that $\psi\cdot\alpha=\sigma$? For example, one can show that the centralizer of the conjugacy classes that are permuted by an outer automorphism are isomorphic; is the converse true? This is consistent with some few cases that I considered, in particular with $Q_8$ and $D_4$, that share the same character table but for which $\Out(Q_4)=S_3$ and $\Out(D_4)=\mathbb{Z}_2$.

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    $\begingroup$ I think it is not common for all symmetries to come from outer automorphisms. If G has exponent n and K is the cyclotomic extension obtained by adding the nth roots of unity to Q, then the Galois group of K over Q acts as symmetries of the character table and these usually don't come from automorphisms. The action is nontrivial as long as there are representations with nonrational characters. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 29, 2022 at 22:23
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    $\begingroup$ The mapping $\psi$ is not always injective as it may happen that $G$ has outer automorphisms which induce the identity permutation on the set of conjugacy classes. The smallest group where this happens has order 32 -- see e.g. here. $\endgroup$
    – Stefan Kohl ♦
    Commented Aug 1, 2022 at 9:09

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