$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^n $ is compact the isometry group $ \Iso(S^n,g) $ is also compact. And every compact group can be realized as the real points of some (reductive) linear algebraic group. Indeed, $ \Iso(S^n,g) = \O_{n+1}(\mathbb{R}) $. The complex points of this group are $ \O_{n+1}(\mathbb{C}) $. And $ \O_{n+1}(\mathbb{C}) $ acts transitively on the tangent bundle of the sphere $ T(S^n) $.

Does this generalize from the round sphere to other compact homogeneous Riemannian manifolds? 

In other words, **Let $ (M,g) $ be a compact Riemannian homogeneous space. Then $ \Iso(M,g) $ is a compact Lie group. So there exists some (reductive) linear algebraic group whose real points are isomorphic to $ \Iso(M,g) $. The question is, does there always exist a linear algebraic group $ G $ such that the real points of $ G $ are isomorphic to the isometry group
$$
G_\mathbb{R} \cong \Iso(M,g)
$$ 
and, in addition, the complex points of $ G $ act (transitively, smoothly) on the tangent bundle $ T(M) $?** 

Note that while the action of $ G_\mathbb{R} $ is by isometries, the action of $ G_\mathbb{C} $ on $ T(M) $ can only be by isometries if $ M $ is parallelizable. So in particular the action of $ \O_{n+1}(\mathbb{C}) $ on $ T(S^n) $ can only be by isometries in the cases $ n=1,3,7 $.

Also note that this result is essentially claimed by Nicolas Tholozan in the comments on https://mathoverflow.net/questions/410443/is-a-quotient-of-real-linear-algebraic-groups-always-a-topological-product-of-co/410467?noredirect=1#comment1058398_410467 I am really just looking for a proof or reference to verify his claim