Finite index of automorphisms that fix normal subgroup does anyone have some thoughts on this:
Let G be a group and K a normal subgroup in G of finite index. Then the set X of all automorphisms phi of G that fix K (meaning phi(K)=K, so not necessarily pointwise) has finite index in Aut G.
Is this even true? I am trying to work out a larger problem and this would be the final step but I am not sure this is even correct.
I tried explaining it by using that the inner automorphisms of G (Int G) are a subset of X (since K is normal). Therefore, if [Aut G:Int G] is finite, so must be [Aut G:X]. But again, I'm not sure this holds...
Thanks very much for your help. I really appreciate it!
 A: If $\alpha$ and $\beta$ are automorphisms of $G$, then the cosets $\alpha X$ and $\beta X$ are the same iff $\alpha(K)=\beta(K)$.  Thus $X$ will fail to have finite index if there are infinitely many different subgroups of $G$ that are conjugate to $K$ under automorphisms of $G$.  For instance, if $G$ is an infinite-dimensional vector space over a finite field and $K$ is subspace such that $G/K$ is finite-dimensional, then automorphisms of $G$ can send $K$ to any subspace $K'$ such that $G/K'$ has the same dimension, and there are infinitely many such subspaces.
A: I don't think this is true without some further finiteness condition that limits the number of images of $K$:
Take $G$ the (two-sided) infinite sequences with entries in 0,1 and as operation component-wise addition (so its an infinite direct product of $C_2$ with itself), and as $K$ the kernel of the projection on one particular component.
Clearly right/left shift are automorphisms of $G$, and under these $K$ has infinitely many images, so the stabilizer of $K$ in the automorphism group must have infinite index.
