In the theory of marginal subgroups there is a following statement:
Suppose $G$ is a group, $M$ is a proper marginal subgroup of $G$. Suppose $\alpha \in {\rm Aut}(G)$ such that $\forall g \in G$ $\alpha(b)b^{-1} \in M$. Then, $|{\rm Stab}_\alpha| > 1$.
Here ${\rm Stab}_\alpha = \{g \in G| \alpha(g) = g\}$.
Proof:
Suppose it were, for a word $w$. Any two elements $x, \alpha(x) \in G$ belong to the same coset, so there are elements $a,b \in M$ such that $x'=ax=xb$. So $w(x_1,\dots,x_n)$ is fixed by $\alpha$ (of $x_1,\dots,x_n$, and hence of $w(x_1,\dots,w_n)$). Hence $w$ can only take values in the stabiliser of $\alpha$ which thus in case when $M$ is proper should be non-trivial, Q.E.D.
My question is:
Can this statement be generalised to something like:
Suppose $G$ is a group, $M$ is a marginal subgroup of $G$. Suppose $\alpha \in {\rm Aut}(G)$ such that $\forall g \in G$ $\alpha(b)b^{-1} \in M$. Then, $|{\rm Stab}_\alpha| > f([G:M])$.
If it can, what $f$ can be taken there?