Surjective orbit tangents give finitely many open orbits

Fix $x$ and write $O=Gx$, $Z=\bar{O}$ with reduced structure. The group $G$ is irreducible: it is the principal open defined by the product of determinants in an affine space, whose polynomial coordinate ring is a domain. Its image and closure are therefore irreducible. By [F1] the orbit is constructible and contains a nonempty open $U$ of $Z$. Every $g\in G$ preserves $Z$, since it preserves $O$ and is a homeomorphism. Transitivity gives $O={\bigcup }_{g\in G}gU$, so $O$ is open in $Z$ and consequently is a locally closed subvariety of $X$.

The orbit map factors through $Z$: on an affine chart every function vanishing on $Z$ pulls back to a function zero at every point of the reduced variety $G$, so it is zero. Apply [F2] to the dominant map $G\to Z$. At some $g$ its differential has rank $\dim Z$, since the image point is regular. The differential of the closed immersion $Z\hookrightarrow X$ is injective, as the map on cotangent spaces is a quotient map. Thus $d_g{f}_x$, considered as mapping into $T_{gx}X$, has rank $\dim Z$. The identity $f_x\circ L_g=(g\cdot -)\circ f_x$ shows that this rank is the same as the rank at the identity, since left translation and the action of $g$ induce tangent isomorphisms. By the hypothesis it follows that $\dim T_{x}X=\dim Z$.

Since $x\in Z\subseteq X$, local dimension obeys $\dim Z\le {\dim }_{x}X\le \dim T_{x}X$; the last inequality is the embedding-dimension bound in [F4]. Step 2.1 makes both inequalities equalities. Hence ${\mathcal{O}}_{X,x}$ is regular and therefore a domain by [F4]. There is exactly one irreducible component $C$ of $X$ through $x$: distinct components through $x$ give distinct minimal primes in this local ring, since localization at $x$ preserves the prime-ideal inclusion correspondence. The irreducible closed subset $Z$ lies in a component through $x$, hence in $C$, and has $\dim Z={\dim }_{x}X=\dim C$. By [F5] it follows that $Z=C$.

Remove from $X$ the finitely many components other than $C$. This gives an open neighbourhood $V$ of $x$ contained in $C$. Since $O$ is open in $C$ by step 1.1, $V\cap O$ is an open neighbourhood of $x$ in $X$ contained in $O$. Translating this neighbourhood by elements of $G$ proves that all of $O$ is open in $X$. All orbits being open, they form a disjoint open cover of the Noetherian, hence quasi-compact, variety $X$. A finite subcover exists by [F3], and disjointness means it lists every nonempty orbit. The empty variety has no orbits; if $G$ is the trivial group the argument says each point is open and quasi-compactness still gives finiteness. AC enters through [F1]–[F5], and no extra infinite choices are made.