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Let $G$ be a group. $\DeclareMathOperator{\Wa}{Wa}\DeclareMathOperator{\Tame}{Tame}\DeclareMathOperator{\Aut}{Aut}$

Let's call $\phi \in \Aut(G)$ verbal automorphism iff $\exists n \in \mathbb{N}, \{a_i\}_{i=0}^n \subset G, \{e_i\}_{i=0}^n \subset \{-1; 1\}$ such that $$\forall t \in G,\;\phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n.$$ One can easily see, that all the verbal automorphisms form a normal subgroup in $\Aut(G)$. Let's denote this subgroup as $\mathrm{Va}(G)$. One can see, that sometimes $\mathrm{Va}(G)$ is a proper subgroup of $\Aut(G)$: for example, $C_2 \times C_2$ has no nontrivial verbal automorphisms, but $\Aut(C_2 \times C_2)$ is isomorphic to $S_3$. Also one can see that a subgroup is invariant under verbal automorphisms iff it is normal.

Let's call $\phi \in \Aut(G)$ inner power automorphism iff it is a composition of an inner automorphism and a universal power automorphism. It is easy to see, that all inner power automorphisms form a normal subgroup in $\Aut(G)$. Let's denote this subgroup as $\mathrm{Ip}(G)$. One can also see that $\mathrm{Ip}(G) \leq \mathrm{Va}(G)$ (as $\forall \phi \in \mathrm{Ip}(G)$, $\exists a \in G, n \in \mathbb{Z}$ such that $\forall t \in G$ $ (\phi(t) = a^{-1}t^na$)).

Is the statement $\mathrm{Va}(G) = \mathrm{Ip}(G)$ always true?

If $G$ is abelian, then it definitely is, as all verbal automorphisms of any abelian group are universal power automorphisms. If $G$ is complete then the statement is also true, as all automorphisms of a complete group are inner. However, I failed to find out anything other than that.

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  • $\begingroup$ Naive question: have you tried when $G$ is a free group (say, with two generators)? $\endgroup$
    – GreginGre
    Jun 22, 2019 at 6:51

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