Let $G$ be a reductive affine algebraic $\mathbb{C}$-group (not necessarily connected). Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$ (again not necessarily connected). Let $x\in X$.

If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed?

NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup, and had not emphasized that I was allowing the adjective "reductive" to include disconnected groups. I have editted the problem to reflect my original intentions.

Let $G$ be a reductive affine algebraic $\mathbb{C}$-group (not necessarily connected). Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$ (again not necessarily connected). Let $x\in X$.

If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed?

NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup, and had not emphasized that I was allowing the adjective "reductive" to include disconnected groups (connected component of identity has trivial unipotent radical). I have editted the problem to reflect my original intentions.

Let $G$ be a reductive affine algebraic $\mathbb{C}$-group. Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$. Let $x\in X$.

If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed  ?

NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup. I have editted the problem to reflect my original intention.

Let $G$ be a reductive affine algebraic $\mathbb{C}$-group (not necessarily connected). Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$ (again not necessarily connected). Let $x\in X$.

If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed?

NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup, and had not emphasized that I was allowing the adjective "reductive" to include disconnected groups. I have editted the problem to reflect my original intentions.