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$\require{AMScd}$For $X$ a (nice enough) topological space, the free loop space $\mathcal{L}X$ is the space of continuous maps from $S^1$ to $X$. This space has a natural $S^1$ action given by rotation of loops, so one can form the homotopy quotient $\mathcal{L}X//S^1$, given, e.g., by the Borel construction $\mathcal{L}BG\times_{S^1} ES^1$.

When $G$ is a finite group, the classifying space $BG$ has a very simple description as a simplicial set (it is the nerve of the one-object groupoid $\mathbf{B}G$ having $G$ as automorphism group of the unique object of $\mathbf{B}G$) and form this one easily get a simplicial description of $\mathcal{L}BG$: it is given by the nerve of the groupoid $G//_{\operatorname{Ad}}G\cong \coprod_{[g]} \mathbf{B}C_G(g)$ where the disjoint union ranges over the conjugacy classes in $G$. Now the question is: which is a simplicial model for $\mathcal{L}BG//S^1$? Whatever this space will be, it will be part of a fiber sequence $$ \mathcal{L}BG \to \mathcal{L}BG//S^1 \to BS^1. $$ And since $BS^1\cong B^2\mathbb{Z}\cong K(\mathbb{Z},2)$, this suggests that whatever a model for $\mathcal{L}BG//S^1$ will be, let us call it $\mathbf{M}$, it will come into a fiber sequence $$ \coprod_{[g]} \mathbf{B}C_G(g) \to \mathbf{M} \to \mathbf{B}^2\mathbb{Z}, $$ where $\mathbf{B}^2\mathbb{Z}$ is the nerve of the 2-groupoid associated to the complex $\mathbf{Z}\to 0$ by the Dold–Kan correspondence (i.e., concretely, the simplicial set whose 2-simplices are decorated by integers, with the obvious cocycle condition). We can look at the various connected components disjointly (after all, $S^1$ is a connected group), so $\mathbf{M}=\coprod_{[g]}\mathbf{M}_{[g]}$ and each $\mathbf{M}_{[g]}$ fits into a fiber sequence $$ \mathbf{B}C_G(g) \to \mathbf{M}_{[g]} \to \mathbf{B}^2\mathbb{Z}. $$ These sequences should not be generally trivial, with the exception of the one involving the identity element $e$ of $G$: in that case $\mathbf{B}C_G(e)=\mathbf{B}G$ corresponds to the space of constant loops in $\mathcal{L}BG$, and on constant loops the $S^1$-action is trivial. So the problem is to produce in a natural way the nontrivial sequences $\mathbf{B}C_G(g) \to \mathbf{M}_{[g]} \to \mathbf{B}^2\mathbb{Z}$ and then show that this construction actually models $\mathcal{L}BG//S^1$. The first part I know how to do, and I'm describing below. But I'm presently not able to verify whether this actually gives a model of $\mathcal{L}BG//S^1$. As the construction is extremely natural one could say “What else could it be?”, but unfortunately this is not yet a proof ….

Here is the construction. For any $g$ in $G$, the cyclic subgroup $\langle g\rangle$ generated by $g$ is a cyclic normal subgroup of the centralizer $C_G(g)$. We therefore have a central extension $$ 1 \to \langle g\rangle \to C_G(g) \to C_G(g)/\langle g\rangle \to 1 $$ and this will be encoded by some cohomology class $[\omega]$ in $H^2(C_G(g)/\langle g\rangle; \langle g\rangle)\cong H^2(C_G(g)/\langle g\rangle; \mathbb{Z}/(k))$, where $k$ is the order of $g$. A 2-cocycle $\omega$ representing $[\omega]$ defines a morphism $$ \omega\colon \mathbf{B}(C_G(g)/\langle g\rangle)\to \mathbf{B}^2(\mathbb{Z}/(k)) $$ whose (homotopy) fiber is $\mathbf{B}(C_G(g))$: we have a fiber sequence $$ \begin{CD} \mathbf{B}C_G(g)@>>> \ast\\ @VVV @VVV\\ \mathbf{B}(C_G(g)/\langle g\rangle)@>\omega>> \mathbf{B}^2(\mathbb{Z}/(k)). \end{CD} $$ We can then define the 2-groupoid $\mathbf{M}_{[g]}$ as the homotopy pullback $$ \begin{CD} \mathbf{M}_{[g]} @>>> \mathbf{B}^2\mathbb{Z}\\ @VVV @VVV\\ \mathbf{B}(C_G(g)/\langle g\rangle) @>\omega>> \mathbf{B}^2(\mathbb{Z}/(k)). \end{CD} $$ The pasting law for (homotopy) pullbacks then gives us the desired fiber sequence $$ \begin{CD} \mathbf{B}C_G(g) @>>> \ast\\ @VVV @VVV \\ \mathbf{M}_{[g]} @>>> \mathbf{B}^2\mathbb{Z}. \end{CD} $$ So everything is as natural as it could. Also, for $g=e$ the defining pullback for $\mathbf{M}_{[e]}$ is $$ \begin{CD} \mathbf{M}_{[e]} @>>> \mathbf{B}^2\mathbb{Z}\\ @VVV @VVV \\ \mathbf{B}G @>\omega>> \quad\mathbf{B}^2(\mathbf{0})\cong \ast. \end{CD} $$ so $\mathbf{M}_{[e]}\cong \mathbf{B}G\times \mathbf{B}^2\mathbb{Z}$, as expected.

The problem is: is this really modelling the right thing?

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