When an action on open dense subvariety by an algebraic group extends to variety A toric variety $X$ over $k$ is a variety which contains an algebraic torus
($T= \mathbb{G}_k^s$)
as a dense open subset such that the action of the torus on itself extends to the whole of
$X$. Slogan: Essentially toric varieties are just fattened tori with an action.
Let $X$ be an algebraic variety which contains a dense open subvariety $U$ and there is an algebraic group $G$ acting on $X$. Are there sufficient conditions when the natural action of $G$ on $U$ extends
to the whole $X$?
The main cases in the scope of my interest are projective varieties
(or weakened to 'proper'). Which role plays the base field $k$ in this extension problem.
Let me also remind that an action by an algebraic group or more general a group scheme $G$ on a algebraic variety
$S$ is a morphism $f: G \times S \to S$ which respects group multiplication morphism
$m: G \times G \to G$. Formally this assumption is equivalent to  these two laws.
Note that's a copy of identical question I asked in MSE a week ago without getting an answer.
 A: The theory of toric embeddings has been generalized to arbitrary homogeneous spaces in the celebrated paper
Luna, D.; Vust, Th. Plongements d'espaces homogènes. (French) [Embeddings of homogeneous spaces] Comment. Math. Helv. 58 (1983), no. 2, 186–245.
Its scope are connected algebraic groups over algebraically closed fields of characteristic zero. Moreover, $U$ is assumed to be a homogeneous variety but that's not really essential.
Of their many reults, one might be of interest for you. Since $U$ is a $G$-variety the Lie algebra $\mathfrak g$ will act by means of vector fields on $U$. If I remember correctly, the following should be true:
Let $G$ be a connected algebraic group (over $\mathbb C$) acting on an irreducible $G$-variety $U$ und let $U\hookrightarrow X$ an open embedding into an irreducible variety $X$. Assume that the $\mathfrak g$-action on $U$ extends to one oon $X$. Then there is open embedding $X\hookrightarrow \overline X$ such that the $G$-action on $U$ extends to a $G$-action on $\overline X$.
Note that if $X$ is complete then $\overline X=X$.
