In this excellent lecture by Matthew Salomone, discussing the correspondence between subgroups and subfields for the polynomial $x^3 - 2$ the following statement is made, which I don't quite get:
[Re: subgroup $\mathbb Z_3=\{e,r,r^2\}$] ... because we are rotating around the $2^{1/3}\text{'s}$ we are leaving the $\zeta_3\text{'s}$ (the cube roots of unity) in place.
Here is the picture on the screen at the time of the statement (above), as well as at the completion of the lattice (below):
Here is the entire diagram with some corrections (clarified in this other post):
My intuition says that the cyclic group permutes the roots of unity, and leaves $\sqrt[3]2$ fixed.
Here's what I learned from the now accepted answer:
The dihedral group $D_3$ leaves the cube roots of unity unchanged with pure rotations, yet they swap when flipping the triangle:
The combined actions of rotation and flipping are better understood with the Cayley diagram in the upper-left hand corner of the image below:
First of all we know that the automorphism of $\mathbb{Q}(\sqrt[3]{2}, \omega_3)$, which fix $\mathbb{Q}$ are determined by the action on $\sqrt[3]{2}$ and $\omega_3$, where $\omega_3$ is a third root of unity. It's trivial to conclude that such an automorphism sends $\sqrt[3]{2}$ to a root of $x^3 - 2$ and $\omega_3$ to a root of $x^2 + x + 1$. Making all possible combinations we get:
$$ i : \begin{array}{lr} \sqrt[3]{2} \to \sqrt[3]{2}\\ \omega_3 \to \omega_3 \end{array} \quad \quad \tau: \begin{array}{lr} \sqrt[3]{2} \to \omega_3\sqrt[3]{2}\\ \omega_3 \to \omega_3 \end{array} \quad \quad \tau^2: \begin{array}{lr} \sqrt[3]{2} \to \omega_3^2\sqrt[3]{2}\\ \omega_3 \to \omega_3 \end{array} $$ $$ \sigma : \begin{array}{lr} \sqrt[3]{2} \to \sqrt[3]{2}\\ \omega_3 \to \omega_3^2 \end{array} \quad \quad \sigma\tau: \begin{array}{lr} \sqrt[3]{2} \to \omega_3^2\sqrt[3]{2}\\ \omega_3 \to \omega_3^2 \end{array} \quad \quad \sigma\tau^2: \begin{array}{lr} \sqrt[3]{2} \to \omega_3\sqrt[3]{2}\\ \omega_3 \to \omega_3^2 \end{array} $$
Now $G = S_3 = D_3$ and the unique cyclic group is $\{i, \tau, \tau^2\}$. Now in the basis $\{1,\sqrt[3]{2},\sqrt[3]{4},\omega_3,\omega_3\sqrt[3]{2},\omega_3\sqrt[3]{4}\}$ of $\mathbb{Q}(\sqrt[3]{2}, \omega_3)$ over $\mathbb{Q}$ the only fixed elements are $1, \omega_3$, which obvioulsy corresponds to the field $\mathbb{Q}(\omega_3)$
Now by checking which elements of the basis remain fixed by a subgroup of $G$, you can determine the corresponding fixed fields.
Although it might not be the right way to think about you can think of a triangle, whose vertices are labeled with the roots of $x^3−2$, the front side labeled with $\omega_3$, while the back with $\omega^2_3$. Obviously the rotations, keep the back and front side the same. Anyway I would prefere explicitly writing the automorphisms, as I did and computing the fixed fields by the subgroups.