Suppose $G=\mathbb{Z}\ast C_n=\langle a, b; b^n\rangle$ is the free product of the infinite cyclic group with a finite cyclic group. Then $G$ has finite outer automorphism group. However, my proofs (and the proofs I have found in the literature*) of this fact use lots of Nielsen-esq automorphisms-of-free-groups theory (with a touch of Whitehead, if I recall correctly).
I was wondering if anyone knows of any short (two-liner), lower-level proofs?
*(Although I will admit I skipped the classic reference, which is a paper of Fouxe-Rabinovitch written in Russian with a MathSciNet entry in German. I speak neither language.)
Are you willing to use the fact that all subgroups of order $n$ in $G$ are conjugate? If so, then we can assume that our automorphism fixes $b$. Then it is enough to prove that it must map $a$ to $b^iab^j$ for some $i,j$, and there are only finitely many such automorphisms. So it is enough to prove that, if $G = \langle g,b \rangle$, then $g = b^iab^j$. If not, then we can assume that the normal form of $g$ has the form $awa$ or $awa^{-1}$ for some nontrivial word $w$ containing $b^{\pm 1}$, and it is not hard to see you cannot generate $a$ from such a $g$ and $b$.