Let $A$ be a (finitely generated) free group and $G$ be a  (finitely generated) free $A$-group - that is, a group with an action of $A$, which is free in the category of groups with an $A$-action. Equivalently, $G$ is isomorphic to the free product of copies of a (finitely generated) free group $H$ (with a free basis $(h_1,...h_n)$) indexed by the elements of $A$, the $A$-action being the obvious one.

I feel that the automorphism group of $G$ (as a group with an $A$-action) should be generated by the automorphisms of H and the automorphisms of the following form:
$h_i\mapsto\negmedspace^ah_i$ (where $a\in A$ and $i\in\{1,\dots,n\}$ are fixed), $h_j\mapsto h_j$ for $j\neq i$.

I think that one can use the usual arguments of Nielsen reduction, adapted in the context of free $A$-groups, to prove this. But the answer should be well-known and written somewhere. Unfortunately, I did not find anything about that by looking for on the web or in classical books of group theory.

Do you have references, or ideas to approach the problem in the most efficient way?

Many thanks in advance.