Let $G$ be a reductive, affine, algebraic group over $\newcommand{\C}{\mathbb C}\C$. Let $X$ be a $G$-variety. For $x\in X$, we write
$$G_x:=\{ g\in G\mid g.x=x\}$$
for its stabilizer and for any subgroup $H\subseteq G$, we write
$$X^H:=\{x\in X\mid H.x=x\}$$
for the $H$-invariants of $X$. We say that $x\in X$ is characterized by its stabilizer if $X^{G_x}=\{x\}$. Let $\{V_\lambda\mid \lambda\in\Lambda\}$ be the irreducible $G$-modules.
Given two points $x,y\in X$, then $x\in\overline{G.y}$ implies $\overline{G.x}\subseteq\overline{G.y}$. Hence, $\C[\overline{G.y}]\twoheadrightarrow\C[\overline{G.x}]$ and thus,
$$\DeclareMathOperator{\mult}{mult}\forall \lambda\in\Lambda:\quad \mult\nolimits_\lambda(\C[\overline{G.x}])\le\mult\nolimits_\lambda(\C[\overline{G.y}])$$
Finding $\lambda\in\Lambda$ violating the above is therefore an "obstruction" for the inclusion of orbit closures.
My question now is the following: If $x$ and $y$ are characterized by their respective stabilizers, does the converse hold? I.e., does the above inequality imply that $x\in\overline{G.y}$? I have been trying to come up with a counterexample, but without success so far.
Intuition: If $G$ acts on a variety $Y$ and $y\in Y$ is characterized by its stabilizer, then you can very easily find counterexamples if you give up the condition that both points are characterized by their respective stabilizers: Consider $X:=Y\times\{z_1,z_2\}$ with $G$ acting trivially on $Z=\{z_1,z_2\}$. Now, the points $x_i:=(y,z_i)$ satisfy $x_1\notin\overline{G.x_2}$ and $\C[\overline{G.x_1}]\cong\C[\overline{G.x_2}]$. In the cases of interest to me, however, both points are characterized by their stabilizer and the question becomes arises whether there are counterexamples under this additional condition.
