Prove that there is no group $G$ s.t. $\operatorname{Aut}(G)=\mathbb{Q}$

Prove that there is no group $G$ s.t. $\operatorname{Aut}(G)=\mathbb{Q}$

I get the feeling that we should proceed by contradiction.

So let $G$ be a group s.t. $\operatorname{Aut}(G)=\mathbb{Q}$. Then we can identify elements of $\mathbb{Q}$ with automorphisms of $G$... and identities such as $\frac{1}{2}*2(g)=1(g)=g$

Can somebody help me find a contradiction?

Suppose $G$ is a group with $\operatorname{Aut}(G)\cong\mathbb{Q}$. If $G$ is abelian, then $f(x)=-x$ is an automorphism of $G$ which satisfies $f^2=1$. But $\mathbb{Q}$ is torsion-free, so this implies $f=1$. But then $G$ is a vector space over $\mathbb{Z}/(2)$, and so its automorphism group is nonabelian if its dimension is greater than $1$ and finite otherwise.

So, $G$ must be nonabelian; say $x,y\in G$ do not commute. Now note that $G$ acts on itself by conjugation, and this gives a homomorphism $\varphi:G\to\operatorname{Aut}(G)\cong\mathbb{Q}$ whose kernel is $Z(G)$, the center of $G$. Note that the subgroup of $\mathbb{Q}$ generated by $\varphi(x)$ and $\varphi(y)$ is cyclic (since every finitely generated subgroup of $\mathbb{Q}$ is cyclic); say it is generated by $\varphi(a)$ for some $a\in G$. Then there are $m,n\in\mathbb{Z}$ and $z,z'\in Z(G)$ such that $x=a^nz$ and $y=a^mz'$. But now we see that $x$ and $y$ actually do commute (since $z$ and $z'$ commute with everything), so we have a contradiction.