$\operatorname{Spaces}/BG$ $\sim$ $\operatorname{Spaces}^G$ $\sim$ $??(\Omega G)$

Question

This is a crosspost (with minor alterations).

For a topological group $G$, assigning to a $G$-space $X$ the (canonical) map $EG\times_GX\to BG$ establishes an equivalence between the homotopy category of $G$-spaces and the homotopy category of spaces over $BG$.

The loop space $\Omega G$ is in turn equivalent to some topological group $A$. Since $G$ is a group, $A$ can be made homotopy commutative (even braided, if I am not mistaken).

Is there some category with the notion of homotopy $\mathscr X(A)$ one may assign to a homotopy commutative topological group $A$ such that when $A$ is homotopy equivalent to the loop space $\Omega G$, then the homotopy category of $\mathscr X(A)$ would be equivalent to the homotopy category of $G$-spaces?

If needed, assume $G$ simply connected or even 2-connected: note that this certainly cannot work for non-connected $G$ since switching to $\Omega G$ loses all information about everything except the connected component of the unity of $G$. For connected non-simply connected $G$, I don't know.

I thought about two possibilities.

First, view a $G$-space $X$ as a continuous homomorphism $G\to\operatorname{Aut}(X)$ to the appropriately topologized group of all self-homeomorphisms of $X$. This gives a map $\Omega G\to\operatorname{aut}_{\operatorname{Aut}(X)}(\operatorname{id}_X)$, the latter being the (homotopy commutative topological) group of self-homotopies of the identity map of $X$. However I do not see a way to recover the action of $G$ on $X$ from this map.

Second, more straightforward and seemingly more promising approach, but somehow I like it less. Note that any $G$-space $X$ gives rise to an action of $\Omega G$ on $\Omega X$. So, one could take for $\mathscr X(A)$ the category of $A$-groups, i. e. topological groups with a continuous action of $A$ by homomorphisms. What confuses me here is that seemingly homotopy commutativity of $A$ drops out. In principle such category exists for any topological group $A$, so why should this work?

Answer 1

If $A$ is a braided ∞-group, the delooping $\def\B{{\sf B}}\B A$ is an ∞-group.

Consider the ∞-category of spaces equipped with an action of the ∞-group $\B A$. Since $\B Ω G≃G$, this ∞-category is equivalent to the ∞-category of $G$-spaces.

The ∞-category of spaces with an action of $\B A$ can be formulated without the functor $\B$. Indeed, there is an adjunction of ∞-categories $\B \dashv Ω$ between braided ∞-groups and ∞-groups. (If we restrict to connected ∞-groups, this adjunction becomes an equivalence.)

An action of $\B A$ on a space $S$ is a homomorphism of ∞-groups $\def\Aut{\mathop{\sf Aut}} \B A → \Aut(S)$, or, equivalently, a homomorphism of braided ∞-groups $A→Ω\Aut(S)$.

Thus, the ∞-category of $G$-spaces can be recovered as the ∞-category of spaces $S$ equipped with a homomorphism of braided ∞-groups $A→Ω\Aut(S)$.

If $A$ is merely a homotopy commutative ∞-group, there is not enough data left to reconstruct the ∞-category of $G$-spaces. Indeed, otherwise we would be able to reconstruct the ∞-group $G$ from the ∞-category of $G$-spaces. But the ∞-group $Ω G$ does not have enough information to reconstruct the ∞-group $G$, unless we equip $Ω G$ with a braiding. Indeed, thanks to the equivalence between ∞-groups and pointed connected spaces, the ∞-group $Ω G$ has exactly the same information as the underlying pointed space of $G$.