If we have $f=x^3-2 \in \mathbb{Q}[x]$ and it has splitting field $L=\mathbb{Q}(\alpha , \omega )$ where $\alpha =\sqrt[3]{2} \ ,\ \omega =\exp{(2\pi i/3)}$.
I know that since $f$ is irreducible and it's roots are $\alpha , \alpha \omega, \alpha \omega^2 $, there is an element $\sigma \in \text{Aut}(L/\mathbb{Q})$ such that $\sigma (\alpha )=\alpha \omega .$
How do I determine the action of $\sigma $ on the other roots?
On
Since $f$ is irreducible and has 3 roots, the Galois group is a subgroup of $S_3$ which is a group of order 6, depending on which permutations of the roots are algebraically allowed. But the real root $\sqrt[3]{2}$ already generates an extension of rank 3 of $\mathbb{Q}$ (it's isomorphic to $\mathbb{Q}[x]/(f)$). Since you still don't have the two non-real roots, you need another extension to split which must now exhaust the rank of $6$ which you have to play with. So the Galois group must be all $S_3$ and all permutations of the roots allowed.
The splitting field is $\mathbb{Q}(\alpha, \sqrt{-3})$ with Galois group $G$ generated by $\sigma : \alpha \to \omega \alpha$ and $\tau : \sqrt{-3} \to -\sqrt{-3}$. Note that $\tau$ acts on $\omega$ because $\omega = (-1 + \sqrt{-3}/2)$.
$\sigma$ is a homomorphism, although its action on the other roots of the polynomial can't be uniquely fixed. All three roots of the polynomial are algebraically 'the same' so there isn't a canonical choice.