Automorphism group of a free product 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 $Aut(G\ast H)$ ?
 A: The known reference is Fuchs-Rabinowitz (or Fouxe-Rabinowitz, Mat. Sbornik, 8 (1940), pp 265-276  and 9 (1941) 183-220. The papers are in Russian and German , see MR0003413 for a review. A more recent reference is McCullough Miller, 
Symmetric automorphisms of free products. Mem. Amer. Math. Soc. 122 (1996), no. 582, viii+97 pp.(MR1329943 )
A: 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 nice paper of Guirardel and Levitt.
