Torsor

Remarks

In the set case, the condition says that the action is free and transitive. In the variety case, $X$ becomes isomorphic to $G$, with $G$ acting by left translation, after a finite extension of $k$. The torsor is trivial if $X \cong G$ over $k$, and this holds if and only if $X$ has a $k$-point.

Example: let $V$ be a vector space and let $A$ be an affine space with translation space $V$. For any two points $a, b \in A$ there is exactly one vector $v$ with $a + v = b$, so $A$ is a $V$-torsor.

Example over a field: let $G = \mu_2 = \set{\pm 1}$ over $\R$ act on $X = \set{x^2 = -1}$ by $x \mapsto \pm x$. Over $\C$, $X$ has the two points $\pm i$, and $G$ swaps them freely. So $X$ is a $\mu_2$-torsor, and it is not trivial, since $X(\R)$ is empty.

Non-example: $\mathbb{G}_m$ acting on $\mathbb{A}^1$ by multiplication is not a torsor. The point $0$ is fixed by every $t$, so the action is neither free nor transitive.