Regular Group Actions and Functions An action of a group $G$ on a set $X$ is regular if for any $x,x'\in X$, there exists a unique $g\in G$ with $gx=x'$.
(a) Give an example.
(b) Let $X$ be a set and let $G$ be a group which acts regularly on $X$. Find all maps $f:X\to X$ such that for every $x\in X$ and $g\in G$, $f(gx)=gf(x)$.
 A: I think these functions are bijective with $X$. Choose some base point $x_0\in X$, and let $f(x_0) = x_0'$. Then if $x\in X$, we have $x = gx_0$ for a unique $g$; the equivariance condition then gives $f(x) = f(gx_0) = gf(x_0) = gx_0'$. So the function $f$ is determined by its value at any element of $X$.
A: Fix $x_0 \in X$ and define a map $G \rightarrow X$ by $g \rightarrow gx_0$, the regular condition implies that this is a bijection, moreover it is $G$ equivariant for the action of $G$ on itself by left multiplication. So from now on I will just assume $X = G$
One obvious source of maps $G \rightarrow G$ that commute with left multiplication are those coming from right multiplication as associativity $g(xh)= (gx)h$ is exactly the $G$ equivariance condition.  I think these should be all such maps but at the moment I can't prove it.
Edit:  after reading rogerl's answer, I see these must be all such maps as there is a unique right multiplication map sending the identity to any given element.
