Let $G$ be a group acting on itself by multiplication by $g \in G$ on the left. I see that this action is free and transitive, thus regular.
Does the converse hold in general? Ie. if a group action is regular then it is multiplication on the left by some $g \in G$?
Let $G$ be a group acting regularly on a set $\Omega$. Choose $i\in \Omega$, and label $i$ with the identity of $G$. Now, for every $\omega\in \Omega$, there is a unique $g\in G$ mapping $i$ to $\omega$. (This is more or less the definition of regular action.) Using this, we get a labelling of $\Omega$ with $G$, and it can easily be checked that what we get is the left-regular action (or right-regular, depending if it's a left or right action in the first place.)