I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:
1) there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and
2) there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.
For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to $F_2$) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?
In general, is there any such automorphism? Is there anything like this in the literature?
