As titled. Let $\omega$ be a primitive cubic root of unity. I find the splitting field $E$ of $x^3-2$ over $F=$ $\mathbb{Q}(\omega)$ is $\mathbb{Q}(\omega,2^{1/3})$. Now I need to determine the elements in $Aut(E/F)$. I know that the automorphism is a subgroup of $S_3$,but I don't know exactly how to determine it. The identity automorphism belongs to $Aut(E/F)$.
If we consider $\sigma(2^{1/3})= \omega\cdot2^{1/3}$, then this is a subgroup which sends $2^{1/3}$ to $\omega\cdot2^{1/3}$, $\omega\cdot2^{1/3}$ to $\omega^2\cdot 2^{1/3}$ and $\omega^2\cdot2^{1/3}$ back to $2^{1/3}$. So this is a permutation of $(123)$ in $S_3$.
Similarly, $\sigma(2^{1/3})= \omega^2\cdot2^{1/3}$ is a permutation of $(132)$. Are these the only automorphism groups? If so, why can't $(13)(2) \ \text{or} \ (12)(3)$ be in $Aut(E/F)$?