Suppose $G$ is a group. $\DeclareMathOperator{\Wa}{Wa}\DeclareMathOperator{\Tame}{Tame}\DeclareMathOperator{\Aut}{Aut}$
Lets 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 $\Aut(G)$. Lets denote this subgroup as $Va(G)$. One can see, that sometimes $Va(G)$ is a proper subgroup: 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.
Lets 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 $\Aut(G)$. Lets denote this subgroup as $Ip(G)$. One can also see that $Ip(G) \leq Va(G)$ (as $\forall \phi \in Ip(G) \exists a \in G, n \in \mathbb{Z}$ such that $\forall t \in G (\phi(t) = a^{-1}t^na$)) and that $Ip(G)$ is always isomorphic to a homomorphic image of $\frac{G}{Z(G)} \times C_{\exp(G)}$, where $\exp(G)$ is the exponent of $G$.
Is the statement $Va(G) = Ip(G)$ 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.
Any help will be appreciated.