Let $\bf Top$ a convenient category of topological spaces, $G$ a group in $\bf Top$, ${}^G\bf Top$ the category of (left) $G$-spaces, and $Sgrp(G)$ the poset of subgroups of $G$.


Define two functors [and prepare yourself to a couple of slight abuses of notation]:


1. ${}^G{\bf Top}\times Sgrp(G)^\text{op}\to {\bf Top}\colon (X,H)\mapsto X^H$, sending a space into the subspace of fixed point of the $H$ action on $X$;
2. $Sgrp(G)\to \mathbf{Top}\colon H\mapsto G/H$ with the induced topology.


Now fix a space $X\in \bf Top$ and consider the functor $Sgrp(G)^\text{op}\times Sgrp(G)\to \bf Top$ defined by $(H,K)\mapsto X^H \times G/K$.


How can you prove (if it is true) that
$$
\int^{H\in Sgrp(G)} X^H\times G/H\cong X?
$$
purely *coendy* proofs are welcome.



A $G$-space as a coend
It doesn't seem to be true. Suppose we take $G = \{-1, 1\}$ with the discrete topology, acting on $X = \mathbb{R}$ by usual multiplication. Here $X^G$ consists of a single point $0$. The coend amounts to a pushout in left $G$-spaces of the diagram 


$$\{-1, 1\} \times \mathbb{R} \stackrel{id \times 0}{\leftarrow} \{-1, 1\} \times 1 \to 1$$ 


and this pushout is preserved by the forgetful functor to $\mathbf{Top}$. It looks like two lines that have been glued together at the origin. 


