Determine the group of Automorphisms of $\mathbb{Q}(\zeta_8)$, where $\zeta_8=\exp(2\pi i/8)$ I've been tasked with the following:

Let $\zeta_8=\exp(2\pi i/8)$. Compute the group of field automorphisms of $\mathbb{Q}(\zeta_8)$.

This seems easy enough, I assume that the problem reduces to showing that automorphisms of $\mathbb{Q}(\zeta_8)$ simply permute the roots of the polynomial $x^7+x^6+\cdots +x+1$, and hence it'll probably be isomorphic to $S_7$ or something similar.
However I've hit a snag: from my knowledge, we only know that a field automorphism $\sigma$ of $\mathbb{Q}(\zeta_8)$ permutes the roots of the cyclotomic polynomial if $\sigma$ fixes $\mathbb{Q}$. I don't quite know how to show this though. We've shown that the only automorphism of $\mathbb{Q}$ is the trivial automorphism, so I assume that ties in somewhere, but I can't see where
Any help is truly appreciated. If you can't tell, this is a homework exercises. So just hints if possible. Thanks in advance.
Edit: is it because we know that $\sigma(\mathbb{Q})\cong\mathbb{Q}/\ker\sigma=\mathbb{Q}$ for each $\sigma\in\text{Aut}\left(\mathbb{Q}(\zeta_8)\right)$? In this case how do we know that $\mathbb{Q}$ doesn't map onto an isomorphic copy of $\mathbb{Q}$ distinct from $\mathbb{Q}$?

Let us start with the basics. Let $f: E\rightarrow E$ be a field isomorphism with $\mathbb{Q}\subset E$ being a subfield. Then we have
$$
f(x+y)=f(x)+f(y); f(xy)=f(x)f(y)
$$
It is now easy to see that 
$$
f(nx)=nf(x), f(1)=1
$$
Therefore we have
$$
f(p*\frac{1}{p})=p*f(\frac{1}{p})=1\rightarrow f(\frac{1}{p})=\frac{1}{p}
$$
and thus $f$ must leave $\mathbb{Q}$ unchanged. If you got stuck with why $f(1)$ must be $1$, think about $f(1^{n})=f(1)^{n}$ for any $n$.  
