**The case of finitely generated ***free* groups

Let me reproduce the proof that Lee Mosher pointed out.

*Claim:*

Let $F=A\ast B$ be a free splitting of a finitely generated free group and $\phi\in Aut(F)$. If $\phi(A)\subseteq A$ then $\phi(A)=A$.

*Proof:*

If $A\ast B$ is a free splitting of $F$ then so is $\phi(A)\ast \phi(B)$. Therefore, $F=\phi(A)\ast \phi(B)$ is the fundamental group of a graph of groups $X$ that consists of a single edge with trivial edge group and vertex groups $\phi(A)$ and $\phi(B)$. Let $T$ be the Bass-Serre covering tree of $X$. By construction, the group $F$ acts on $T$ with trivial edge stabilizers and with vertex stabilizers conjugate to $\phi(A)$ and $\phi(B)$ respectively. In particular, there exists a vertex $v\in V(T)$ such that $F_v=\phi(A)$, where $F_v$ denotes the point stabilizer of $v$ under the action of $F$.

The subgroup $A$ of $F$ also acts on the tree $T$ and the edge stabilizers under this action are also trivial. Our assumption $\phi(A)\subseteq A$ implies that the vertex stabilizer $A_v=F_v\cap A=\phi(A)\cap A$ is in fact given by $\phi(A)$. Now choose a fundamental domain for the action of $A$ on $T$ accordingly such that the quotient graph of groups $Y$ with underlying topological graph $A\backslash T$ has $\phi(A)$ as one of its vertex groups. Bass-Serre theory tells us that $A$ is the fundamental group of $Y$, and since $Y$ has trivial edge groups it follows that $A$ contains $\phi(A)$ as a free factor.

We haven't yet made use of the fact that we are dealing with *free* groups ($A$ and $B$ are free, as all subgroups of free groups are free). Since $\phi$ restricted to $A$ is an isomorphism onto its image $\phi(A)$, the two free groups $A$ and $\phi(A)$ have the same finite rank. Therefore, $A$ cannot contain $\phi(A)$ as a *proper* free factor and hence $\phi(A)=A$.

**Does this statement also hold for ***arbitrary* finitely generated groups?

No, not in general.

In '*A Finitely Related Group with An Isomorphic Proper Factor Group*', J. London Math. Soc. (1951) s1-26(1): 59-61, Graham Higman gives an example of a finitely presented group $G$ that is isomorphic to a proper free factor of itself. In other words, there exists a free splitting $G=G_1\ast G_2$ and an isomorphism $$\phi\colon G_1\ast G_2\stackrel{\cong}{\longrightarrow} G_1.$$ The free product of $\phi$ with its inverse $\phi^{-1}$ is then an automorphism $$\Phi=\phi\ast\phi^{-1}\colon (G_1\ast G_2)\ast {G_1}'\stackrel{\cong}{\longrightarrow} G_1\ast ({G_1}'\ast G_2)$$ which has the property that $\Phi(G_1\ast G_2)=G_1\subseteq G_1\ast G_2$, but equality does not hold.