Is $Y/K$ homeomorphic to $Y'$ as defined below -

Question

Let $G$ be a topological group acting on a topological space $X$ in such a way that there are only finitely many orbits. We will fix points $x_1,\cdots,x_n\in X$ and let $X=\bigcup_{i=1}^n G\cdot x_i$ . Let $G_i=\text{ Stab }(x_i)$.

Suppose $H$ is another topological group with a surjective morphism $\phi:H\to G$. Consider the space $H\times X$ and define an equivalence on it by $(h,x)\sim (h',x')$ iff $x=x'$ and $\phi(h(h')^{-1})\in G_i$ where $G\cdot x_i$ is the unique orbit containing $x$. Let $Y:=(H\times X)/\sim$

Let $K=\ker\phi$. Let $K$ act on $H\times X$ by $k\cdot(h,x)=(kh,x)$. Then this action descends to an action on $Y$.

Let $Y':=(H/K\times X)/\sim'$ where $(Kh,x)\sim'(Kh',x')$ iff $x=x'$ and $\bar\phi(Kh(h')^{-1})\in G_i$ where $\bar\phi:H/K\to G$ is the isomorphism induced by $\phi$.

Question - Is $Y/K$ homeomorphic to $Y'$?

Intuitively I think it is true but I am struggling to write the steps.

1. Define a map $\tilde F:Y\to Y'$ given by $[(h,x)]\mapsto[(Kh,x)]$. This is clearly well defined and continuous and surjective.

2. Check that the map descends to a map $F:Y/K\to Y'$. Let $[(h',x)]=[(kh,x)]$. Then $F[(h',x)]=F[(kh,x)]=[(Kkh,x)]=[(Kh,x)]=F[(h,x)]$. So the map descends and by universal property of quotient spaces is continuous. Also this map is clearly surjective.

3. Injectivity of $F$ - I'm stuck with this.

4. $F$ is an open map - I know that the quotient map $Y\to Y/K$ is open so if $\tilde F$ is open then so is $F$. But I am stuck there as well.

Thank you.