Question about Enriched Categories and Functors How would one describe the process of enriching a category C over some monoidal category D? Is there some functor between them that adds structure to the hom-sets? 
 A: While Paul's answer is correct, it is also possible to start with an ordinary category $C$ and "enrich it" by adding structure stuff.  One sort of tautological way to do this is to specify a $V$-enriched category $\hat{C}$ whose underlying ordinary category (in the sense mentioned by Todd) is isomorphic (or equivalent) to $C$.  A more interesting way is to specify a 2-variable adjunction $V\times C\to C$ with some natural "associativity" conditions (see e.g. Def. 14.3 here); this yields not just any enrichment of $C$ but one that is tensored and cotensored.
A: The question is mis-conceived.
(In any other disciplines, much less naive questions than this are closed rapidly and dismissively, but I am opposed to this practice.)
An enriched category is not a category to which extra structure has been added (and certainly not by a functor), indeed it need not be a category at all.
A $\mathbb{V}$-enriched category $\mathcal C$ has a collection of objects, just as an ordinary category does, though for clarity I am going to call them vertices.  However, instead of a set ${\mathcal C}(X,Y)$ of morphisms between any two vertices, it has an object of them, where this object belongs to the enriching category $\mathbb V$. As nLab or numerous textbooks will explain, $\mathbb V$ has to be a monoidal category, whose tensor product and unit provide composition and identities in $\mathcal C$.
Probably you are imagining that $\mathbb V$ is a category of vector spaces or topological spaces or something else that has "underlying sets".
Forget that and think of something else, like this:
Let ${\mathbb V}={\mathbb R}^+$, where $a\to b$ iff $a\geq b$, the tensor product is $+$ and the unit is $0$.
Please work out the characterisation of this example as a valuable exercise for yourself before you look it up in this famous paper.
