Is there an example of a finitely generated subgroup $\Gamma \subset \mathrm{GL}_n(\mathbb{C})$ such that the group of outer automorphisms $\mathrm{Out}(\Gamma)$ contains finite subgroups of unbounded orders?
1 Answer
Yes. Furthermore, every recursively presented countable group embeds in such a group. Indeed, first Higman's embedding reduces to proving that every finitely presented group embeds into such a group (if you want all finite groups, it is enough to use a single known f.p. group with all finite groups in it, e.g., Thompson's group $V$).
Now the Rips construction says that for every f.p. group $Q$, there exists a $C'(1/6)$ f.p. small cancelation group $G$ and a finitely generated and normal subgroup $N$ such that $G/N\simeq Q$. Hence $Q$ embeds into $\mathrm{Out}(N)$.
Finally, we use HaglundWiseAgol's results, which ensure that every $C'(1/6)$ f.p. small cancelation group is linear over $\mathbf{Z}$.
Edit: as suggested by HJWR in a comment, it is more direct to directly use the HaglundWise "Rips construction" (GAFA 2008other link), which directly provides a group that is linear over $\mathbf{Z}$, without need of the Wise machinery plus Agol.

4$\begingroup$ For the record: to get this for all $C'(1/6)$ groups uses the full strength of the AgolWise machinery, which is pretty challenging. But HaglundWise gave a virtually special (in particular, linear over $\mathbb{Z}$) version of the Rips construction in their GAFA paper `Special cube complexes'. This is much easier. $\endgroup$– HJRWMay 30 at 9:33