The symmetric group $\Sigma_k$ acts on $X=F({\mathbb R}^n,k)$, the ordered configuration space of $k$ points in ${\mathbb R}^n$. If $n$ is odd, the cohomology $H^*(X;{\mathbb Q})$ is a rank-one free module over ${\mathbb Q}[\Sigma_k]$.
This is checked by calculating the character of this representation. (Note also that this is obvious for $n=1$.)
Are there known (one-vector) bases of this module?
This isn't exactly an answer to your question, but here's how I like to think about the fact that you quoted.
Let's assume for a minute that $n=2$, so that we can think of $X$ as the complement of the braid arrangement in $\mathbb{C}^k$. Let $G = \mathbb{Z}/2\mathbb{Z}$, which acts on $X$ by complex conjugation.
Replace $\mathbb{Q}$ with a field $F$ of characteristic $2$. The $G$-equivariant cohomology ring $H^*_G(X; F)$ is a free module over $H^*_G(pt; F) \cong F[x]$ with the property that specializing at $x=0$ gives $H^* (X; F)$ and specializing at $x=1$ gives $H^*(X^G;F)$.
Thus we have a family of $\Sigma_k$ representations over the $F$-affine line interpolating between $H^* (X; F)$ and $H^*(X^G; F)$. Since the category of $\Sigma_k$ representations is semisimple, these two representations have to be isomorphic. The fact that $H^*(X^G; F)$ is the regular representation is obvious.
This is a good way to see that $H^*(X; F)$ is isomorphic to the regular representation of $\Sigma_k$. I'm not sure how to modify this argument to get $H^* (X; \mathbb{Q})$. I'm also not sure if this will help you find a cyclic vector, since it does not give you an explicit isomorphism between $H^* (X; F)$ and $H^* (X^G; F)$.
By the way, for $n>2$ you can do something similar, where $G$ acts on $\mathbb{R}^n$ by negating the last $n-1$ coordinates.
