Suppose that $G$ is a finite group scheme over a field $k$ (we may want to assume that $k$ is perfect). How does one tell whether there exists a free action of $G$ on the function field $k(t)$ in one variable? By this I mean that there exists an action $G \times_{\mathop{\rm Spec}k}\mathop{\rm Spec}k(t) \to \mathop{\rm Spec}k(t)$, making $\mathop{\rm Spec}k(t)$ into a $G$-torsor over a scheme (necessarily of the form $\mathop{\rm Spec} k(s)$, where $s \in k(t)$, by Lüroth's theorem).

The question is a very natural one when one studies essential dimension of group schemes: see http://www.math.ubc.ca/~reichst/lens-notes6-27-8.pdf for a nice survey of the topic of essential dimension, and http://front.math.ucdavis.edu/1001.3988 for the essential dimension of group schemes. When $G$ is smooth over $k$, then it is easy to see that the action extends to an action on $\mathbb{P}^1$, so $G$ must be a subgroup of ${\rm PGL}_{2,k}$; but when $G$ is not smooth it is not all clear to us that this must happen. The sheaf of automorphisms of $k(t)$ over $k$ is enormous in positive characteristic, and we find it very hard to see what group schemes it contains.

For example, how about twisted forms of the group scheme $\mu_p$, where $p$ is the characteristic of the field? I would conjecture that most of them can't act freely on $k(t)$, but we can't prove it.

finitebase $G$, by looking at geometric fibers over $G$ your setup amounts to an action of $G$ on a dense open of the projective line (as for ordinary finite groups). But then it perhaps get complicated, since automorphism functor of such opens is generally not representable. – BCnrd Apr 5 '10 at 16:32