From Abstract Algebra by Charles Pinter:
Let $K$ be the root field of some polynomial over $F$. The number of automorphisms of $K$ fixing $F$ is equal to the degree of $K$ over $F$.
Let $[K:F]=n$, so we have $n$ distinct roots. From the discussion in the chapter we know that such automorphisms permute the roots of the polynomial and hence completely determined by the permutations. However, we know that there're $n!$ permutations of the $n$ roots, which contradicts the theorem.
Later in the chapter the author notes that
Not every permutation of the roots of $a(x)$ need be in the Galois group of $a(x)$.
and he gives an example
Since $x^4-10x^2+1$ has four roots, there're $4!=24$ permutations of its roots, only four of which are in its Galois group. This is because only four of the permutations are genuine symmetries of $x^4-10x^2+1$, in the sense that they determine automorphisms of the root field.
which doesn't really make sense to me, since he never defined genuine symmetries or determining automorphisms. I'm guessing that most of the automorphisms can be compositions of others, but even so why are we not counting them?
The Galois group of the extension $K/F$ is the group of automorphisms of $K$ fixing $F$, i.e., $\alpha : K \to K$ such that $\alpha(1) = 1$, $\alpha(xy) = \alpha(x)\alpha(y)$, $\alpha(x+y) = \alpha(x) + \alpha(y)$, and such that $\alpha(x) = x$ for all $x \in F$.
Suppose $K$ is generated by the roots of a polynomial $f(x)$. Then for any automorphism $\alpha$ we have $f(x) = 0 \implies f(\alpha(x)) = \alpha(f(x)) = 0$, so $\alpha$ determines a permutation $\pi_\alpha$ of the roots of $f$. If $R$ is the set of roots of $f$ we therefore have a homomorphism $\mathrm{Gal}(K/F) \to \mathrm{Sym}(R)$, and it is injective because the roots generate $K$ (if $\alpha$ fixes the roots $x \in R$ then it fixes everything), but it is not generally surjective.
In your example $x^4 - 10x^2 + 1$ the roots are actually $t,-t, 1/t, -1/t$, where $t = \sqrt{5 + 2 \sqrt 6}$. Any automorphism $\alpha$ is determined by where it sends $t$, since $\alpha(-t) = -\alpha(t)$, $\alpha(1/t) = 1/\alpha(t)$, and $\alpha(-1/t) = -1/\alpha(t)$. Therefore only $4$ of the $4!$ permutations of $\{t, -t, 1/t, -1/t\}$ are possible.