Conjugacy classes as left Kan extension of forgetful functor Let $\mathbf{Set}$, $\mathbf{Grp}$, and $\mathbf{Grp}^{\rm conj}$ denote the categories of sets and functions, groups and homomorphisms, and groups and homomorphisms up to conjugation, respectively. (Thus the fundamental group is a functor from path-connected spaces to the latter.)
Let $X$ be the forgetful functor from $\mathbf{Grp}$ to $\mathbf{Set}$, $A$ the obvious functor from $\mathbf{Grp}$ to $\mathbf{Grp}^{\rm conj}$, and $L$ the functor from $\mathbf{Grp}^{\rm conj}$ to $\mathbf{Set}$ sending a group to its set of conjugacy classes. Then is $L$ the left Kan extension of $X$ along $A$?
 A: An alternative to Maxime's sophisticated proof is to observe that $X$ and $L$ are both corepresented by the free group, because a group homomorphism from a free group (up to conjugacy) is the same as choosing an element (up to conjugacy).
That is, $X = \hom_{\mathrm{Grp}}(\mathbb{Z},-)$ and $L = \hom_{\mathrm{Grp}^{\mathrm{conj}}}(A(\mathbb{Z}),-)$.
In general, the left Kan extension of a (co)representable functor is the (co)representable functor by the image of the representing object (this is the naturality of the Yoneda embedding).
See nLab for example.
Putting the two together, we get the required equivalence:
$$
A_!X
= A_!\hom_{\mathrm{Grp}}(\mathbb{Z},-)
= \hom_{\mathrm{Grp}^\mathrm{conj}}(A(\mathbb{Z}),-)
= L
$$
A: $\newcommand{\Gp}{\mathbf{Grp}}
\newcommand{\conj}{^{\mathbf{conj}}} 
\newcommand{\Gpd}{\mathbf{Gpd}}
\newcommand{\Set}{\mathbf{Set}}
\newcommand{\ho}{\mathrm{ho}}
\newcommand{\Fun}{\mathrm{Fun}}
\newcommand{\Lan}{\mathrm{Lan}}
\newcommand{\colim}{\mathrm{colim}}
$
Yes. There is a "by-hands proof", where you just compute the colimits that are involved, but there turns out to be a nice proof if you go to $(2,1)$-categories. Namely, $\Gp\conj$ is equivalent to the homotopy category of a natural $(2,1)$-category: its objects are groups, its $1$-morphisms are group morphisms, and its $2$-morphisms are conjugations : if $f,g: G\to Q$ are morphisms, a $2$-morphism $f\implies g$ is an element $q\in Q$ such that $f^q = g$ (where $f^q : qf(-)q^{-1}$).
I'm going to call this $\Gpd$, and it's relatively easy to observe that $\Gp \simeq \Gpd_{1/}$ in a way that the functor $A$ can be identified with the composite $\Gp\simeq \Gpd_{1/}\to \Gpd\to \ho(\Gpd)\simeq \Gp\conj$.
Because $\Set$ is a $1$-category, restriction along $\Gpd\to\ho(\Gpd)$ induces an equivalence $\Fun(\ho(\Gpd),\Set)\to \Fun(\Gpd,\Set)$, so that left Kan extension is "silly": it's the observation that the functor factors through $\ho$.
Therefore, it suffices to prove that the left Kan extension of $\Gpd_{1/}\to \Set$ along $\Gpd_{1/}\to\Gpd$ is given by the set of conjugacy classes. But now we can reap the benefits of this approach: $\Gpd_{1/}\to\Gpd$ is a left fibration (as is $C_{x/}\to C$ for any $(\infty,1)$-category $C$ and object $x$), so that the left Kan extension of a functor $f:\Gpd_{1/}\to \Set$ along $p:\Gpd_{1/}\to \Gpd$ is given at $G$ by the colimit over the fiber of $f$ (note that $\Gp\to \Gp\conj$ is not a left fibration).
More precisely: $\Lan_p f(G) = \colim_{\alpha \in p^{-1}(G)} f(1\xrightarrow{\alpha} G)$. Here, $f(1\to G)$ is just the underlying set of $G$, and the $1$-groupoid $p^{-1}(G)$ is equivalent to $BG$, and it acts on $G=f(1\xrightarrow{\alpha} G) $ by conjugacy, so that this colimit is the quotient of $G$ by its conjugacy action.
So $\Lan_p X$ is the functor from $\Gpd$ to $\Set$ that sends $G$ to its set of conjugacy classes, and this factors through $\ho(\Gpd)\simeq \Gp\conj$ in the appropriate way.
