TITLE: Almost pointwise inner automorphism of free products of groups.
QUESTION [5 upvotes]: Let $A,B$ be groups, let $G = A\ast B$ be their free product and let $\phi \in \text{Aut}(G)$ be a automorphism of $G$. We say that $\phi$ is pointwise inner if $\phi(g) \sim_G g$ (there is $w \in G$ such that $\phi(g) = wgw^{-1}$) for every $g \in G$. It is not too difficult to show that if $\phi$ is pointwise inner then $\phi$ is actually inner.
However, what if we assume that $\phi(g) \sim_G g$ for all $g \in G$ such that $|g|=2$, where $|g|$ denotes the length of $g$? Can one then show that $\phi(g)\sim_G g$ for all $g\in G$ such that $|g|=1$?

REPLY [2 votes]: I think the answer is no.
Let $G$ be a free group generated by $a, b$. Then $G=\langle a \rangle * \langle b\rangle$. Let $\phi: a\rightarrow b, b\rightarrow a$. Then $\phi$ is an automorphism of $G$. Let $g\in G$ with $|g|=2$, then $g=a_1b_1$ where $a_1\in \langle a \rangle$ and $b_1\in \langle b \rangle$. Now $\phi(g)=b_1a_1\sim g$. But $\phi(a)=b \not \sim a$ by Theorem 4.6 of W. Magnus, A. Karrass, D. Solitar, Combinatorial Group Theory, Pure Appl. Math., vol. XIII, Wiley–Interscience, New York.
------------OK, this above answer is not true. But its method works---
Let $a, b$ are order $2$ and $G=\langle a \rangle * \langle b\rangle$. Let $\phi: a\rightarrow b, b\rightarrow a$. Then the above works.