Let $G$ be a reductive group acting on an affine variety $X$. Let $x\in X$ such that the stabiliser of $x$ in $G$ is a subgroup $H$ which is itself a reductive group. Then, it is easy to see that any point in the orbit $Gx$ is stabilised by some conjugate of the group $H$ in $G$. Now, as the orbit $Gx$ isn’t necessarily closed, we can consider an arbitrary point $y$ in the closure of this orbit. Is the point $y$ necessarily stabilised by some conjugate of $H$? To simplify things, we can assume that the orbit $Gy$ of $y$ is closed in $X$, and then, does the question have a positive answer in this case?
As a reality check, this statement is trivially true when $H=G$ or if $H$ is trivial. The question is essentially asking if being closed by a conjugate of $H$ is an algebraic condition, but I don’t know how to see it.
See Proposition 4.19 here for when $y$ has a closed orbit: https://web.northeastern.edu/iloseu/Jose_S16.pdf.
The idea of the proof is to use Luna's slice theorem. Assuming that $y$ has a closed orbit, there exists a slice $S$ through $y$ and a map: $$\phi:G \times^{G_y} S \rightarrow X,$$ given by $(g,s)\mapsto gs$. The theorem tells us, among other things, that the image contains an open neighbourhood of $y$. Then, for any $y'$ in the neighbourhood of $y$, we compute the stabiliser $G_{y'}$ by computing the stabiliser of a point in its pre-image under $\phi$, say $(g,s)$.
Then, it is a straightforward computation to check that $G_{y'}=G_{(g,s)} = g(G_y)_sg^{-1}\subseteq gG_yg^{-1},$ or $g^{-1}G_{y'}g \subseteq G_y$. As this is true for all points $y'$ in a neighbourhood of $y$, the claim follows.
When the orbit isn’t closed, one can repeat the argument with a slight modification. Consider $X \setminus (\overline{Gy} \setminus Gy)$. Then, we see that $Gy$ is closed in this subvariety and still lies in the closure of $Gx$, and so, we can repeat the argument above.