Suppose that $G$ and $H$ are groups (not isomorphic) and $G\ast H$ the free product. Let $Aut(G)$, $Aut(H)$ be the automorphism groups of $G$ and $H$. What is $Aut(G\ast H)$ ?



Automorphism group of a free product
This is not an answer, but it's too long for a comment and makes an important point that I hope is useful.


It's not enough to think of your group just as some free product. To understand its automorphism group you need to think of it as a free product in a *canonical* way, and for this you need the Grushko decomposition. Recall:



> 
> **Grushko's decomposition theorem:** If $G$ is finitely generated then
> 
> 
> $G=G\_1\*\ldots\* G\_n\*F\_k$
> 
> 
> where each $G\_i$ is freely indecomposable (ie isn't $\mathbb{Z}$ and doesn't split as a free product) and $F\_k$ is free of rank $k$. Furthermore, the $G\_i$ are determined up to conjugacy (and reordering), and $k$ is determined.
> 
> 
> 


The key thing to notice here is that the $F\_k$ factor is *not* determined up to conjugacy, and therefore if $k>0$ then there are complicated automorphisms that arise essentially from the different possible ways to choose $F\_k$. However, the picture is relatively simple if $k=0$.


A modern reference on the (outer) automorphism groups of free products is [this](http://arxiv.org/abs/math/0501288) nice paper of Guirardel and Levitt.


