Right Regular Representation by Inverse is Transitive Group Action

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $*: G \times G \to G$ be the group action:
 * $\forall g, h \in G: g * h = \rho_{g^{-1}} \left({h}\right)$

where $\rho_g$ is the right regular representation of $G$ with respect to $g$.

Then $*$ is a transitive group action.

Proof
Let $g, h \in G$.

Then:

$h$ is arbitrary, therefore the above holds for all $h \in G$.

By definition of orbit:
 * $\operatorname{Orb} \left({h}\right) = \left\{{t \in G: \exists g \in G: g * h = t}\right\}$

That is:
 * $\operatorname{Orb} \left({h}\right) = G$

Hence the result by definition of transitive group action.