It can be shown the Galois group of $x^3 - x - 1$ is $S_3$. What is the orbit or $x$ under this permutation action? In other words, we can find three things, elements of the extension field that are literally permutated by Galois group?
Certainly we can name the roots $x=x_1, x_2, x_3$ and say they are permuted. So I guess I'm asking to express $x_2$ and $x_3$ as polynomials in $x_1=x$. The group $S_3$ is acting on the $\mathbb{Q}$-vector space:
$$K = \mathbb{Q}[x]/(x^3 - x - 1) \simeq \mathbb{Q}\cdot 1 + \mathbb{Q}\cdot x + \mathbb{Q}\cdot x^2 $$
and therefore we are getting a representation of the permutation group.
Edit The Galois closure $L$ is itself a quadratic extension of $K$, with $[L:\mathbb{Q}]=[L:K][K:\mathbb{Q}]=6$.
There is a way to save all of this. $S_3$ acts on $L = \mathbb{Q}[x]/p(x)$ for some irreducible sextic polynomial $p(x)$ with $\deg p = 6$.
In remains to find $p(x)$ and then... what is the orbit of $x$ under $S_3$ in the splitting field of $K$ ?
Another example: the cubic $x^3 - 3x - 1$ has Galois group is $A_3$ (which has only $|A_3|=3$ elements) and acts on the extension $\mathbb{Q}[x]/(x^3 - x - 1)$ by two permutations: $x \mapsto x^2 - x - 2$ and $x \mapsto -x^2 + 2$, and we get a representation.
No, the Galois group is not acting on $K=\Bbb Q[x]/(x^3-x-1)$ since that isn't a Galois extension of $\Bbb Q$. Rather it acts on the Galois closure which is a degree $6$ extension. One cannot express $x_2$ as a polynomial in $x_1=x$, but of course $x_3=-x_1-x_2$.