Nonexistence of a functor from $Group$ to $Set$ taking each group to its set of automorphism

I am struggling with this question: show that there does not exist functor from $Group$ to $Set$ taking each group to its set of automorphisms. I have thought about it for a while now, not having any insight.

There is a thread on MSE showing the nonexistence of a functor taking each $\mathcal{G}$ in $Group$ to Aut($\mathcal{G}$) in $Group$, which is relevant but not of help.

Edit: I know that for some 1-1 homomorphism $f:G\longrightarrow H$, there is an automorphism $A$ on $G$ such that there is more than one automorphisms $g_1$ and $g_2$ on $H$, for which $g_1\circ f=g_2\circ f=f\circ A$.

On the other hand, for some 1-1 homomorphism $f:G\longrightarrow H$, there is an automorphism $A$ on $G$, such that there is no automorphism $g$ on $H$ for which $g\circ f=f\circ A$

This rules out one way of defining the functor, but I am not sure how it helps in the general case.

I have thought of one way of doing this, but there might be easier ways. Let $\mathcal{F}$ be a functor with the properties you describe.
Suppose that we have groups $G$, $H$ with homomorphisms $\phi:G \to H$ and $\psi:H \to G$ with $\psi \phi = {\rm Id}_G$. Then $\mathcal{F}(\psi\phi) = \mathcal{F}(\psi) \mathcal{F}(\phi)$ is the identity map on $\mathcal{F}(G)$. So if we can find such an example with $|{\rm Aut}(H)| < |{\rm Aut(G)}|$, then this is not possible, and we have a contradiction. Note that, in this situation, $H = K \rtimes G$ with $K = \ker(\psi)$.
One such example is $G=C_2^n$ elementary abelian of order $2^n$ and $H = S_3^n$. Then ${\rm Aut}(G) = {\rm GL}(n,2)$ and ${\rm Aut}(H) = S_3 \wr S_n$. We have $|{\rm Aut}(H)| < |{\rm Aut(G)}|$ for $n \ge 5$.
• Thanks for the answer. Could you please explain what $H=S^3_n$ is? – Chase Mar 8 '14 at 17:09
• It is the direct product of $n$ copies of the symmetric group $S_3$ (or the dihedral group $D_6$ if you prefer). It's not too hard to show that its automorphism group must permute these $n$ direct factors, so we get $|{\rm Aut}(H)| = 6^nn!$. – Derek Holt Mar 8 '14 at 17:59