A functor $\mathcal G\to \mathbf{Set}$ is the same as a left $G$-set I'm trying to understand the first part of Example 1.2.8 from here: https://arxiv.org/pdf/1612.09375.pdf
Let $Ob(\mathcal G)=\{\star\}$. A functor $F:\mathcal G\to \mathbf{Set}$ consists of:


*

*An assignment $F: Ob(\mathcal G)\to Ob(\mathbf{Set}),\star\mapsto S_\star$. This is indeed "the same as" choosing a set (I guess formally this means that the class of such assignments is in bijection with the class of sets.)

*An assignment $F: \mathcal G(\star,\star)\to\mathbf {Set}(S_\star,S_\star)$ satisfying $F(f\circ g)=F(f)\circ F(g)$ and $F(1_\star)=1_{S_\star}$ for all $f,g:\star\to\star$. Since $\mathcal G(\star,\star)$ is bijective to the set of elements of the monoid $G$ and since $\circ$ in the category corresponds to $\cdot$ in the monoid, the above can be written as $F:G\to \mathbf {Set}(\star,\star)$ subject to $F(f\cdot g)=F(f)\circ F(g)$, $F(1_\star)=1_{S_\star}$.


How does one get from the above that $F:\mathcal G\to\mathbf{Set}$ consists of a set $S$ together with, for each $g\in G$, a function $F(g):S\to S$, satisfying the functoriality axioms, as claimed in the text linked above?
 A: As Daniel says in the comments, the claim is nothing more than 'unpacking' the definition of functor in this particular case. 
The first realization one has to have is that a groupoid $\mathcal{G}$ that has only one object $*$ "is a group". That is, the arrows $G = \mathcal{G}(*,*)$ for a group and determine $\mathcal{G}$ (recall that for any category one could forget the objects and just work with arrows, as the former are represented by identities).
Now, to be formal, consider the category $G\mathsf{Set}$ of $G$-sets toghether to functions that commute with the $G$-actions. We can think of the objects here as pairs $(X,\rho)$ where $\rho : G \to S(X)$ is the action. 
Now, as per your bullet points we can define the functor
$$
\begin{align}
\mathcal{\Gamma} :\mathsf{Set}&^\mathcal{G} \to G\mathsf{Set}\\
& F \longmapsto (F* , \rho_F) \\
& \downarrow_{\eta}\ \mapsto \quad \downarrow_{\eta_*}\\
& F' \mapsto (F'*,\rho_{F'})
\end{align}
$$
where $\rho_F(g)(x) = F(g)(x)$ and $\eta_* : F* \to F'*$ is the $*$-component of the natural transformation $\eta$.
You can check that this is not only an equivalence of categories but a category isomorphism, with the inverse sending $(X,\rho)$ to the functor that maps $* \mapsto X$ and $ * \xrightarrow{g} * $ to $\rho(g) : X \to X$. Likewise, a $G$-function $h$ from $(X,\rho)$ to $(X',\rho')$ gives rise to a natural transformation whose only component is $h$ itself.
A: Let $M$ be a monoid regarded as a one-object category $\mathscr M$ with unique object $\star$.
We first show that any functor $F: \mathscr M\to\mathbf{Set}$ gives rise to a left $M$-set (which is, by definition, a pair $(S,\cdot)$, where $S$ is a set and $\cdot$ is a left action of $M$, i.e., a map $$M\times S\to M,\\(m,s)\mapsto m\cdot s $$ such that $(m_1m_2)\cdot s=m_1\cdot(m_2\cdot s)$ and $e\cdot s=s$, where $e$ is the identity of $M$.)
Let $S=F(\star)$ and define the map $M\times S\to S$, written $(m,s)\mapsto m\cdot s$, by $m\cdot s=F(m)(s)$. (Here we identify the elements of $M$ with the arrows of $\mathscr M$ and use one and the same letter $m$ to denote them.) We need to check that the axioms of action hold. Well, since $F$ is a functor, we have $F(1_\star)=1_S$ and $F(m_1\circ m_2)=F(m_1)\circ F(m_2)$. Evaluating both sides of each equation at $s\in S$, we get, respectively, $F(1_\star)(s)=1_S(s)$ and $(m_1\circ m_2)(s)=F(m_1)(F(m_2)(s))$ or, equivalently, $1_\star\cdot s=s$ and $(m_1\circ m_2)\cdot s=m_1\cdot (m_2\cdot s) $. Since $\circ$ corresponds to multiplication in $M$ and $1_\star$ corresponds to $e$, this translates to $e\cdot s=s$ and $(m_1m_2)\cdot s=m_1\cdot(m_2\cdot s)$. In this way, $F$ gives rise to a left $M$-set.
Conversely, consider a left $M$-set $(S,\cdot)$. Define the functor $F:\mathscr M\to \mathbf{Set}$ as follows. Define the image of the unique object $\star$ by $F(\star)=S$. If $m:\star\to \star$ is an arrow in $\mathscr M$, define $F(m): S\to S$ by $F(m)(s)=m\cdot s$. Let us prove functoriality: $$F(m_1\circ m_2)=(m_1\circ m_2)\cdot s=(m_1m_2)\cdot s=m_1\cdot (m_2\cdot s)=\\ m_1\cdot F(m_2)(s)=F(m_1)(F(m_2)(s))=(F(m_1)\circ F(m_2))(s).$$ The second requirement for $F$ being a functor is checked similarly. This shows that to every left $M$-set there corresponds a functor $\mathscr M\to \mathbf{Set}$.
