The base field is the field of complex numbers. Let $G$ be a connected linear algebraic group. Let $X$ be an almost homogeneous algebraic variety, i.e. $G$ acts on $X$ with a dense open orbit $U \simeq G/H$. Then it is well-known that $\mathrm{Aut}^G(U) \simeq N_G(H)/H$, where $N_G(H)$ is the normalizer of the algebraic subgroup $H$ in $G$, is naturally endowed with an algebraic group structure. On the other hand, $\mathrm{Aut}^G(X)$ is an abstract subgroup of $\mathrm{Aut}^G(U)$. Here I denote by $\mathrm{Aut}^G(X)$ the group of $G$-equivariant automorphisms of the variety $X$.
My question is: How to show that $\mathrm{Aut}^G(X)$ is in fact an algebraic subgroup of $\mathrm{Aut}^G(U)$?
