The Automorphism Group $\Gamma(\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q})$ is trivial if $n$ is odd.

The $\mathbb Q$-automorphisms of $F_n$ are completely determined by where they send $\sqrt[n]2$, and they must send it to another $n$-th root of $2$. But if $n$ is odd, then no other $n$-th root of $2$ is in $F_n$, and if $n$ is even, then only one other $n$-th root of $2$ is in $F_n$ (you might need to prove these facts beforehand). So there are only one/two possible $\mathbb Q$-automorphisms.