Let $P\in \mathbb{Q}[X]$ a polynomial and $\alpha_1, ..., \alpha_n \in \mathbb{C}$ its distinct complex roots and suppose its splitting field is $\mathbb{Q}(\alpha_1)$.
Then an automorphism of $\mathbb{Q}(\alpha_1)$ leaves $\mathbb{Q}$ invariant so it must send a root of $P$ to another root of $P$, and it is fully determined by the image of $\alpha_1$ so there are at most $n$ such automorphism. But are there exactly $n$ automorphisms of $\mathbb{Q}(\alpha_1)$ ?
I'm asking this because was looking at the polynomial $P=X^4-2X^2+9$ and one can compute that its roots are $\sqrt2 +i, \sqrt2 -i, -\sqrt2 +i, -\sqrt2 -i$, its splitting field is $\mathbb{Q}(\sqrt2 +i)$, and I wanted to determine all the automorphisms of $\mathbb{Q}(\sqrt2 +i)$.
Given a Galois extension $K/L$ for fields $K$ and $L$, the size of the automorphism group of $K$ over $L$ is equal to $[K:L]$. And if $K$ is constructed by adjoining a single element $\alpha$ to $L$, then this number is also equal to the degree of the minimal polynomial of $\alpha$.
Since this is a Galois extension, and thus a separable extension, the degree of the minimal polynomial of $\alpha$ (which in your case might or might not be $P$) is equal to the number of distinct roots of that minimal polynomial. So yes, in your case there are exactly $n$ automorphisms, as $\Bbb Q(\alpha_1)$ being the splitting field of $P$ means that all roots of $P$ are roots of the minimal polynomial of $\alpha_1$.