Let $G$ be a group, $M$ a set and $\alpha: G \times M \to M$ a group action. For $g \in G$, let $\operatorname{Fix}_{\alpha}(g) = \{ m \in M \mid g \cdot_{\alpha} m = m \}$.
It is easily seen that if $g, h \in G$, then $\operatorname{Fix}_{\alpha}(h g h^{-1}) = h \operatorname{Fix}_{\alpha}(g)$, so in particular $|\operatorname{Fix}_{\alpha}(h g h^{-1})| = |\operatorname{Fix}_{\alpha}(g)|$.
Now let $\varphi \in \operatorname{Aut}(G)$. Does it hold that $|\operatorname{Fix}_{\alpha}(\varphi(g))| = |\operatorname{Fix}_{\alpha}(g)|$?
To expand on @lulu's comment:
Consider a nontrivial action $\alpha: G \times M \to M$, meaning that $\exists g_0 \in G, m_0 \in M : g_0 \cdot_{\alpha} m_0 \neq m_0$. Now let $\beta: (G \times G) \times M \to M, (g_1, g_2) \cdot_{\beta} m = g_2 \cdot_{\alpha} m$. Then we have $\operatorname{Fix}_{\beta}(g_0, e) = M$, but $\operatorname{Fix}_{\beta}(e, g_0) \neq M$ because $m_0 \notin \operatorname{Fix}_{\beta}(e, g_0)$. However, $(g_0, e)$ and $(e, g_0)$ are related by the automorphism of $G \times G$ swapping the two factors.