So I'm working through Galois Theory by Ian Stewart, and I'm on on chapters 10/11, which discuss automorphisms, galois groups, etc. There is an example 10.7 that I'm trying to work through:
"Let $K = \mathbb{Q}(\xi)$ where $\xi = e^{2\pi i / 5} \in \mathbb{C}$. Now $\xi^5 = 1$ and $\mathbb{Q}(\xi)$ consists of all elements $p + q\xi + r\xi^2 + s\xi^3 + t\xi^4$ where $p, q, r, s, t \in \mathbb{Q}$. The Galois group of $\mathbb{Q}(\xi) : \mathbb{Q}$ is easy to find, for if $\alpha$ is a $\mathbb{Q}$-automorphism of $\mathbb{Q}(\xi)$ then $(\alpha (\xi))^5 = \alpha (\xi^5) = \alpha(1) = 1$, so that $\alpha (\xi) = \xi, \xi^2, \xi^3, \xi^4$. This gives four candidates for $\mathbb{Q}$-automorphisms:
$\alpha_1 : p + q\xi + r\xi^2 + s\xi^3 + t\xi^4 \rightarrow p + q\xi + r\xi^2 + s\xi^3 + t\xi^4$
$\alpha_2 : p + q\xi + r\xi^2 + s\xi^3 + t\xi^4 \rightarrow p + s\xi + q\xi^2 + t\xi^3 + r\xi^4$
$\alpha_3 : p + q\xi + r\xi^2 + s\xi^3 + t\xi^4 \rightarrow p + r\xi + t\xi^2 + q\xi^3 + s\xi^4$
$\alpha_4 : p + q\xi + r\xi^2 + s\xi^3 + t\xi^4 \rightarrow p + t\xi + s\xi^2 + r\xi^3 + q\xi^4$
It is easy to check that all of these are $\mathbb{Q}$-automorphisms..."
I started working through this example because problem 10.3 says to work through this example with $\xi = e^{2\pi i / 7}$, and I think I understand everything up until the actual automorphisms. It seems like the automorphisms are just moving coefficients around, but I can't tell how, since it's not every possible permutation of the coefficients. How exactly are these four automorphisms found?
Since $\alpha$ is an automorphism, we have $\alpha (\xi^k)= \alpha(\xi)^k$, so given $\alpha(\xi)$, the other follows.
In your example $\alpha_2$ sends $\xi$ to $\xi^2$ and you can see then $\xi^2$ is sent to $\xi^4$, $\xi^3$ to $\xi^6=\xi$ and $\xi^4$ to $\xi^8=\xi^3$