Is a monoid $M \in \bf{\text{Mon}}$ equivalent to the same monoid $M \in \bf{\text{Cat}}$? We usually say that a monoid is just a category with only one object. Thus, for every monoid $M$, viewed in as a category, there corresponds a monoid $M$ in the category of monoids and vice-versa. But this a little bit vague, isn't? What machinery do we need to make this more elegant?
Is there a way to I establish that $M \in \bf{\text{Mon}}$ and $M \in \bf{\text{Cat}}$ are equivalent? My guesses:
($\Rightarrow$) I think we can define a functor $F : \bf{\text{Mon}} \to \bf{\text{Cat}}$ that takes objects to themselves and each monoid homomorphisms to a functor between those monoids viewed as categories.
($\Leftarrow$) Really have no idea. What domain would this functor have?
Also, since $\bf{\text{Mon}}$ and $\bf{\text{Cat}}$ are different categories, is an isomorphism even possible?
Thanks!
 A: The more precise and invariant statement is that the category of monoids is equivalent to the category of pointed connected categories; "connected" means there is a single isomorphism class of object (as opposed to just a single object; that's an evil condition), and "pointed" means you've distinguished such an object. 
If you remove "pointed," then connected categories naturally form a 2-category, not a category, and this 2-category is a bit more complicated than the 1-category of monoids. 
A: You want to establish something like an equivalence of categories between monoids and the full subcategory of categories on the categories with a single object. Given a monoid $M$, there is a category $FM$, but you shouldn't say this is "the object $M$ itself" before you've proved your proposition! $FM$ is instead simply defined to be the single-object category whose morphisms $FM(*,*)$ are given by $M$. Then $F$ is a functor-just check explicitly that the monoid homomorphism axioms are the same as the axioms for a functor between one-object categories.
The inverse is similarly simple: send a one-object category to the endomorphisms of its single object. This is a monoid, and you'll check that these two functors are inverse to each other.
