How to interpret topologically that the equalizer in Groupoids of ${\rm id}, {\rm id}: BG \rightrightarrows BG$ is $G/G$ (adjoint action)? Let $G$ be a (discrete) group, and $1/G$ the corresponding groupoid with one object.  Consider the diagram in (the 2-category) Groupoids with one vertex, labeled $1/G$, the one arrow from that vertex to itself, given by the identity map.
$$ \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix}$$
(This diagram is equivalent to the pair of parallel arrows $1/G \overset{\rm id}{\underset{\rm id}\rightrightarrows} 1/G$.  Note that I am not filling in the loop with a 2-cell.)
A cute fact is that the ("2-") limit of this diagram in Groupoids is the action groupoid $G/G$ of the adjoint action of $G$ on itself.  (See e.g. 2 limit in nLab or HTT Chapter 4 for a definition of limits.)
Now, in homotopological terms, the groupoid $1/G$ looks like the classifying space ${\rm B}G$, and the above diagram looks like ${\rm B}G \times S^1$.  I have the possibly-mistaken impression that limits are supposed to look like topological cones (but maybe this is because we use words like "cone" when talking about limits).
Question: In terms of homotopy, how should I visualize the limit cone
$$ \lim\left( \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} \right) \quad \begin{matrix} {\huge \to} \\ {\large \circlearrowleft \!\!\!\!\!\! \circlearrowleft} \end{matrix} \quad \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} $$
?
(Edits: per Quid's request in the comments, I replaced some broken images with diagrams, trying to reconstruct them from memory.  $\circlearrowleft \!\!\!\!\! \circlearrowleft$ is my attempt at a doubled circle arrow, i.e. a 2-cell filling in the cone walls.)
 A: You should think of the limit of that diagram as "loops in $BG$."  A loop in $BG$ is a principal $G$-bundle on $S^1$.  Every principal $G$-bundle on the circle comes from taking the trivial principal $G$-bundle on $[0,1]$ and identifying the fibers over $0$ and $1$ (making this the basepoint). If you like left principal bundles, this gluing map has to be right multiplication by an element $g$ of $G$.  Of course, you can still do gauge transformations on the circle, and these will have the effect of conjugating $g$ by the value of the gauge transformation at the basepoint.
Thus, principal bundles on $S^1$ can be thought of as $G/G$.
A: To slightly amplify what Ben wrote, the diagram is precisely a presentation of $Map(S^1,BG)=L(BG)$ rather than of $BG\times S^1$. More generally the loop space of a space $X$ can be presented as the homotopy fiber product $LX= X\times_{X\times X} X$, the self-intersection of the diagonal, which is a slightly different way (which I find more convenient) to present self-homotopies of the identity map of $X$. In the case of a groupoid (or a stack) this results in the inertia groupoid, i.e., objects (points) together with automorphisms. Again in the case of $BG$ we have one object (the trivial $G$-torsor on a point, in one presentation) and its automorphisms form a $G$, with automorphisms given by $G$ acting adjointly. 
On the level of functions/chains (interpretation depending on your context), rather than points, you get a formula that looks more like what you wrote, i.e. $$F(X) \otimes S^1= F(X) \otimes_{F(X)\otimes F(X)} F(X),$$ aka the Hochschild homology (or chains) of functions on $X$.
