Let $G$ be a complex reductive group acting algebraically on a complex affine variety $X$. Is it true that an orbit $G.x$ is closed in $X$ if and only if the stabilizer $G(x)$ of $x$ is a reductive group?
I would appreciate references or ideas on how to prove it.
No, this is not true. Consider $\mathbb C^\times$ acting on $\mathbb C$ by multiplication. The stabilizer of $1\in\mathbb C$ in $\mathbb C^\times$ is equal to $\{1\}$ (which is reductive), the orbit is equal to $\mathbb C^\times$, but the closure of this orbit is $\mathbb C$.
Also, take any nonreductive group $G$ acting trivially on $X$, then the orbit of any point $x\in X$ is the closed set $\{x\}$, but the stabilizer is the nonreductive group $G$.