I apologize in advance if this is naive.
In this answer Conjugation in a groupoid it is said that given a groupoid $\mathcal G$, and an arbitrary function $\mu:\mathcal G_0\to \bigcup_{x\in \mathcal G_0}\mathcal G(x,x)$ s.t. $\mu_x\in \mathcal G(x,x)$ we can define an endofunctor $F_\mu\in[\mathcal G,\mathcal G]$ by $F_\mu(g)=\mu^{-1}_yg\mu_x$.
What I find interesting is that that functor $F_\mu$, induced by $\mu$ a arbitrary choice function, turns $\mu$ into a natural transformation of the same functor it defines. Let $g:x\to y$ we have $\mu_yF_\mu(g)=g\mu_x$.
$$\mu:F_\mu\Rightarrow {\bf 1}_{\mathcal G}$$
$Q1$ - So does this phenomenon has a name? What is the big picture?
$Q2$ - For example what are the most general conditions on $\mathcal C$ for which a choice function $\sigma:\mathcal C_0\to \bigcup_{x\in \mathcal C_0}\mathcal C(x,x)$ induces a functor that turn $\sigma$ into one of its natural transformations?
I wonder if some deep structure or functor lies under this mechanism like, say, a "functor-ish something" $\sigma\mapsto (F_\sigma,\sigma) \in[\mathcal C,\mathcal C]/{\bf 1}_{\mathcal C}$ to the slice category or something like that.
$Q3$ - This last is a bit more vague. To what extent, given an endofunctor $F$, can we find a special natural transformation $\phi_F:F\Rightarrow {\bf 1}_{\mathcal G}$ that "has the most information possible" about $F$ in a similar or weaker way to how $\mu$ "knows everything/define" about $F_\mu$?
Motivation
I exapand my comment to @MartinBrandenburg. The main motivation I have in mind is that of groupoids (like the linked answer) but it comes to my mind a possible generalization to the case, say, of the non-canonical choices of isomorphisms $X\sim|X|$ in sets or $V\simeq k^{{\rm dim} V}$ for vector spaces. Where from a choice of isomorphisms we define functors that makes the choice natural for the functor defined. In those cases in fact we can define a functor $$| - |:{\bf Set}\to {\bf Set} \quad\textit{ or }\quad k^{{\rm dim} -}:{\bf FinVec}_k\to {\bf FinVec}_k$$ only if previously we are given a collection of bijections $$i_X:|X|\to X \quad\textit{ and }\quad \varphi_V:k^{{\rm dim} V}\to V.$$
Only after we have that we can define the component on arrows and get functors. And only when we have functoriality $i$ and $\varphi$ become natural transformations.
This construction works in generality and what you are effectively doing is giving a description of all endofunctors naturally isomorphic to the identity functor. The basic (general) point is that if $F,G\in[\mathcal{C},\mathcal{C}]$ are endofunctors and $(\mu_x)_{x\in\mathcal{C}}$ is a natural isomorphism $F\Rightarrow G$, then the commutativity of the following diagram for all $x,y\in\mathcal{C}$ and $g\colon x\rightarrow y$ \begin{equation} \require{AMScd} \begin{CD} F(x) @>{\mu_x}>> G(x)\\ @VV{F(g)}V @VV{G(g)}V\\ F(y) @>{\mu_y}>> G(y) \end{CD} \end{equation} already determines $F$, namely $F(g)=\mu_y^{-1}\circ G(g)\circ\mu_x$ (note the similarity to your formula).
Here's a formulation of the phenomenon involving the slice category that encompasses the examples described in your comments. If $F\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})$ is a function and $\mu_x\colon F(x)\rightarrow x$ an isomorphism for each $x\in\mathcal{C}$, then there is one and only one functor $F$ acting as $F$ on objects and for which $(\mu_x)_{x\in\mathcal{C}}$ defines a natural isomorphism $F\Rightarrow1_{\mathcal{C}}$. This functor acts on a morphism $g\colon x\rightarrow y$ as $F(g)=\mu_y^{-1}\circ g\circ\mu_x$ (compare above). Confirming that this is a functor is easily checked. This furnishes an injective map $\coprod_{F\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)\rightarrow[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$. In fact, note that if you have two objects $\mu\colon F\Rightarrow1_{\mathcal{C}}$ and $\mu^{\prime}\colon F^{\prime}\Rightarrow1_{\mathcal{C}}$ in $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$ such that $\mu$ and $\mu^{\prime}$ are isomorphisms, then there is one and only one morphism $\mu\rightarrow\mu^{\prime}$ in $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$ (which is necessarily an isomorphism). Thus, if we interpret the domain of this map as an indiscrete category, it defines an embedding of this category in $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$ whose image is precisely the full subcategory on the endofunctors with a natural isomorphism to $1_{\mathcal{C}}$ (it is in thise sense that you are describing endofunctors naturally isomorphic to the identity functor).
As far as Q3 is concerned, if you look back at the diagram, you can notice that $F$ is in fact already determined by $\mu$ and $G$ if each $\mu_x,x\in\mathcal{C}$ is a monomorphism. So your question is sort of like "what's the next best thing to a monomorphism?". I do not have a good answer for that question.