TITLE: Stabilizer in automorphism group of free group of a certain finite-index subgroup
QUESTION [6 upvotes]: The following question arose in my research.  I'd be interested in an answer to it, but I'd also be interested in general techniques for solving this kind of problem (or, even better, pointers to computer programs that can solve it automatically).
Consider the group $G = \text{Aut}(F_2)$.  Of course, $G$ acts on $F_2$.  Let $x$ and $y$ be the generators for $F_2$, and let $K \subset F_2$ be the normal closure of the set $\{x^4, x^2 y^{-2}, y^{-1} x y x\}$.  It is standard that $F_2/K$ is the $8$-element group of quaternions.  
Question : What are generators for the subgroup $\{\text{$g \in G$ $|$ $g(K) = K$}\}$ of $G$?

REPLY [10 votes]: Here's a short proof by hand that $K$ is characteristic in $F_2$.
The natural map $Q_8\to V=\mathbb{Z}/2\oplus\mathbb{Z}/2$ induces $\eta:F_2\to H_1(F_2,\mathbb{Z}/2)\cong V$.  The kernel $N=\ker\eta$ is therefore characteristic and contains $K$ with index two.
Suppose now that $\phi$ is an automorphism of $F_2$ that does not preserve $K$.  Then $\phi$ descends to an automorphism of $V$ which does not lift to an automorphism of $Q_8$.  But it's easy to check that every automorphism of $V$ does, in fact, lift.