I wonder whether it is true that *the composition of two GIT-quotients is another GIT-quotient*. It should be an analogue of a set-theoretic formula $X/(G \times H)\simeq (X/G)/H$ but with GIT-quotients instead.

**Background. Mumford's theorem (GIT, thm. 1.10).** Let $G$ be a reductive group acting on an algebraic variety $X$ and (linearly) on a vector space $V$, and let $\phi: X \to \mathbb PV$ be a $G$-morphism. Then there is a non-empty open subset
$$X_{ss,\phi}=\{ x \in X : \text{for any $v\in V$ representing $\phi(x)\in\mathbb PV$}$$
$$\text{the closure $\overline{Gv}$ of the $G$-orbit of $v$ does not contain $0\in V$ } \}$$
such that there exists a ~~geometric quotient~~ categorical quotient $p: X_{ss,\phi} \to X_{ss,\phi}/G$. It is called (semistable) GIT-quotient.

**Problem.** Let $G \times H$ be a reductive group acting on an algebraic variety $X$ and (linearly) on a vector space $V$, and let $\phi: X \to \mathbb PV$ be a $G$-morphism. By Mumford's theorem, there is the GIT-quotient $p_1: X_{ss, \phi} \to X_{ss, \phi} / G$ and it satisfies a universal of categorical quotient, $H$ also acts on $Y:=X_{ss, \phi} / G$.

Now let $H$ act (linearly) on a vector space $W$ and $\psi: Y \to \mathbb PW$ be an $H$-morphism, so that there is a GIT-quotient $p_2: Y_{ss,\psi} \to Y_{ss,\psi} / H$. So let us denote $Z:=Y_{ss,\psi} / H$.

Now suppose that not only $H$ but $G \times H$ acts (linearly) on $W$. Let $\chi: X \to \mathbb PW$ be a $G \times H$-morphism such that the following diagram is commutative: $$\begin{array}{ccc} X & \xrightarrow{\chi} & \mathbb PW \\ \uparrow i && \uparrow \psi \\ X_{ss,\phi} & \xrightarrow{p} &Y \end{array}$$

So there is a GIT-quotient $X_{ss, \chi} \to X_{ss, \chi} / (G \times H)$. I wonder whether $$X_{ss, \chi}=X_{ss, \phi} \cap p_1^{-1}(Y_{ss, \psi})$$ and $$X_{ss, \chi} / (G \times H) \simeq Z.$$

Any suggestions: references, counterexamples or corrections are welcome!