Functor from a group considered as a category with one object to the category of sets I am learning category theory as a hobby. In the book I am studying from, I was looking at an example of a functor. I want to understand what this functor looks like.
If $G$ is a group, a functor $F: G \to \mathbf{Set}$ picks out a set $A = F(\star)$, together with a homomorphism from $G$ to the group of permutations of $A$. This is a permutation representation of $G$. 
I am confused about some things here:


*

*What is $\star$, is it the single object in the group considered as a category?

*Why is this homomorphism from $G$ to the group of permutations of $A$ coming into the picture, doesn't a functor just need to map objects to some objects and morphisms to some morphisms? 

*The book says if that a group can be considered as a category consisting of the single object $G$ where all the morphisms are isomorphisms. Why is this the case? What does this category look like?
 A: Think about transformation groups.  You have a space $X$ with symmetries $\operatorname{Aut} X$, and a transformation group is a subgroup $G\subset\operatorname{Aut} X$.  Elements of $G$ are automorphisms $X\to X$.
You can think of an abstract group as being a transformation group for an abstract object $\star$.  Elements of $G$ are automorphisms $\star \to \star$.  You are not meant to think of $\star$ as being anything other than a thing that exists.  You can think of the one-object category as being the group, or you can think of $G$ as being $\operatorname{Aut}(\star)$. You could say that both points of view are the group, with the kind of metonymy that mathematicians tend to use.
A homomorphism $G\to \operatorname{Perm}(A)$ (with $A$ a set and $\operatorname{Perm}(A)$ the set of all bijections $A\to A$) can be thought of as a functor $F$ that


*

*sends $\star$ to $A$, and

*sends an element $g\in G$, thought of as an automorphism $g:\star\to\star$, to the permutation $F(g):A\to A$.


$F$ being a functor means that $F(gh)=F(g)F(h)$ and $F(1_G)=\operatorname{id}_A$, so $F$ restricted to the morphisms of $G$'s category is the homomorphism.
A one-object category is called a monoid.  Call the object $\star$.  Recall that there is an identity map $1:\star\to\star$ in the category.
If every $f:\star\to\star$ is an isomorphism (i.e., there is an inverse $g:\star\to\star$ such that $gf=1$ and $fg=1$), then $\operatorname{End}(\star)$ satisfies all the axioms of a group.
A: *

*Yes. It is common to use $*$ as the name of the object in a one-object category; more generally, it is often used as the name of the unique object contained in a singleton set. ($\star$ is less common, but is surely what's meant)

*A functor must additionally preserve products of morphisms, as well as send identity morphisms to identity morphisms.

*Because if you write what "one object category whose morphisms are all isomorphisms" means, you'll see its almost verbatim the same thing as what "a monoid in which every element has an inverse" means (plus the statement that there's one object).

