It is well-known that if a reduced algebraic group $G$ acts on a *separated reduced* scheme $X$, and $G$ acts trivially on a dense open subscheme $U\subseteq X$, then the action is trivial.

If $X$ is non-reduced, the standard counterexample is $\def\AA{\mathbb A}$$\AA^1$ with an embedded point, with the group acting non-trivially on the embedded point. If $X$ is non-separated, the standard counterexample is $\AA^1$ with a doubled origin with $G$ swapping the two origins. My question is whether the separated hypothesis can be removed if $G$ is assumed to be *connected*.

Suppose $G$ is a

connectedgroup scheme acting on areducedscheme $X$, and that $G$ acts trivially on a dense open subscheme $U\subseteq X$. Must the action be trivial?

For reference, the basic argument for the original version is that the graphs of the two morphisms $G\times X\to X$ (projection and action) are closed subschemes of $G\times X\times X$ (since the graphs are pullbacks of the diagonal of $X$, which is assumed separated) which share a common dense open (the image of $G\times U$). Since $G\times X\times X$ is reduced, this means that the two graphs agree, so the action and projection maps are the same.

^{†}~~For finite-characteristic people, I actually assume $G$ is reduced, but this is fine since the action on a reduced scheme must factor through the reduction of $G$ anyway.~~