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I think this will work. There are a finite number of orbits of the action of $G$ on $X$ precisely when there is an open orbit and a finite number of orbits on the complement of the orbit. Hence, it is enough to show that if there is an open orbit of a point over the smaller field precisely when there is an open orbit over the larger field. One direction is clear so it is enough to show that there is an open orbit over the larger field, there is one over the smaller field. However, consider the closed subscheme $S:=\{(g,x)\in G\times X | gx=x\}$ and the projection $S\to X$ on the second variable. A point of $X$ has an open orbit precisely when the fibre has the smallest possible dimension (when $X$ is irreducible which we may assume). However, there is an open subset (defined over the smaller field) of $X$ with fibres of minimal dimension.

Addendum: As for the bijection between the orbits this is proved the same way, we have an open orbit which gives one orbit over each field and then we use Noetherian induction.

2 deleted 3 characters in body

I think this will work. There are a finite number of orbits of the action of $G$ on $X$ precisely when there is an open orbit and a finite number of orbits on the complement of the orbit. Hence, it is enough to show that if there is an open orbit of a point over the smaller field precisely when there is an open orbit over the larger field. One direction is clear so it is enough to show that if there is an open orbit over the larger field, there is one over the smaller field. However, consider the closed subscheme $S:=\{(g,x)\in G\times X | gx=x\}$ and the projection $S\to X$ on the second variable. A point of $X$ has an open orbit precisely when the fibre has the smallest possible dimension (when $X$ is irreducible which we may assume). However, there is an open subset (defined over the smaller field) of $X$ with fibres of minimal dimension.

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I think this will work. There are a finite number of orbits of the action of $G$ on $X$ precisely when there is an open orbit and a finite number of orbits on the complement of the orbit. Hence, it is enough to show that if there is an open orbit of a point over the smaller field precisely when there is an open orbit over the larger field. One direction is clear so it is enough to show that if there is an open orbit over the larger field, there is one over the smaller field. However, consider the closed subscheme $S:=\{(g,x)\in G\times X | gx=x\}$ and the projection $S\to X$ on the second variable. A point of $X$ has an open orbit precisely when the fibre has the smallest possible dimension (when $X$ is irreducible which we may assume). However, there is an open subset (defined over the smaller field) of $X$ with fibres of minimal dimension.