I'm struggling to understand this. Let $\mathbb{Q}(\xi)$ be the splitting field for $x^n - 1$, so here $\xi$ is the primitive n-th root of unity. If I consider the possible automorphisms of this field, I will get:
$$\iota \textrm{ (the identity)}, \qquad \textrm{and} \qquad \sigma_k: \xi \mapsto \xi^k, \quad2 \leq k< n.$$
I will have $n-1$ elements, same as $Z_{n-1}$. How can it be shown that this is actually $Z_n^{\times}$ instead? What am I missing?
Simply, not all the maps $\sigma_k : \xi \mapsto \xi^k$ are field automorphisms. For example, consider $n = 4$; in this case, we usually denote the primitive root by $i$. Then, $\sigma_2$ maps both $\pm i$ to $-1$, so $\sigma_k$ is not an automorphism of $\Bbb Q[i]$ fixing $\Bbb Q$.
On the other hand, any automorphism must permute the roots of unity, so the map $\xi \mapsto \xi^k$ must be invertible if it is to be an automorphism, but this only holds if $k, n$ are coprime, that is, if $k \in \Bbb Z_n^{\times}$. Note that $\Bbb Z_n^{\times} \cong \Bbb Z_{n - 1}$ iff $k, n$ are coprime for all $1 \leq k < n$, that is, iff $n = 1$ or $n$ prime.
Again for $n = 4$ we have $\Bbb Z_4^{\times} = \{1, 3\}$, so the only two automorphisms are characterized by $i \mapsto i$ (the identity) and $i \mapsto i^3 = -i$ (complex conjugation).