Invariant neighborhood in a G-CW complex Let $G$ be a discrete group and $X$ be a $G$-CW complex. For any $x\in X$ and open neighborhood $U$ of $x$,  I am interested in the question that whether we can find a $G_x$-invariant open neighborhood $V$ of $x$ that is contained in $U$.  This is not true in general, for example, if $X$ is the cone over $\mathbb R$ with the $G=\mathbb Z$ action by translation, then the cone point $x_0$ has a neighborhood which does not contain any $G_{x_0}$-invariant open neighborhood. But I am wondering whether the following is true:
Any $G$-CW complex $X$ is $G$-homotopy equivalent to a $G$-CW complex $Y$ so that any open neighborhood $U$ of $y\in Y$ contains a $G_y$-invariant open neighborhood $V$.
As a special case, I am wondering whether the following is true:
Let $\mathcal F$ be a family of subgroups of $G$ which is closed under conjugation and finite intersection. Can we always find a model $E_{\mathcal F}G$ for the universal $G$-CW complex relative to $\mathcal F$ so that any open neighborhood of $x\in E_{\mathcal F}G$ contains a $G_x$-invariant open neighborhood?
Thank you!
 A: I will call the property of a $G$-CW-complex that inside every neighborhood of a point one can find a $G$-invariant neighborhood property $A$.
As in your example, a graph where an edge stabilizer has infinite index in one of the adjacent vertex stabilizers does not have property $A$.
Further $G$-subcomplexes of $G$-CW-complexes with property $A$ have property $A$; given an open set in the subcomplex, extend it to an open set of the whole complex, choose that invariant neighborhood there and intersect back.
Now look at the free group in $a,b$ with the family of subgroups $\mathcal{F}$ containing the trivial group and all conjugates of $\langle a \rangle$ and $\langle b \rangle$. One model for $E_FG$ is given by the Bass-Serre tree, which does not have property $A$. But why can't there be a better model?
Then its one-skeleton would also have property $A$. But by the defining property of $E_FG$ we can find a point $p_a$ with $\langle a \rangle$ is contained in the stabilizer of $p_a$. Since $\langle a \rangle$ is maximal in $F$, it actually follows that $\langle a \rangle$ is the stabilizer of $p_a$. Choose $p_b$ analogously.
Since the one-skeleton is connected, we can choose a finite path from $p_a$ to $p_b$. we want to show that on that path there is some edge, whose stabilizer group has infinite index in the stabilizer of one adjacent vertex. If this was not the case, then $\langle a \rangle$ and $\langle b \rangle$ would be comensurable in the free group, which they are not.
