If $M$ is a monoidal category, an enriched category over $M$ is a category $C$ whose hom-sets are viewed as objects in $M$. And a monoidal category $M$ is said to be closed if the tensor product functor has a right adjoint, known as the internal Hom functor.
My question is, are “monoidal category enriched over itself” and “closed monoidal category” equivalent? To put it another way, do all monoidal categories with internal Hom sets have internal Hom functors, and do all monoidal categories with internal Hom functors have internal Hom sets?
If not, does anyone know of counterexamples in either direction?
The two concepts are distinct.
In general it is true that every monoidal closed category is enriched over itself, and this fact is exploited for studying enriched presheaves.
The converse does not hold as the following example will show.
Consider the category $\mathbf {Top}$ of topological spaces and continuous maps between them, with the cartesian monoidal structure. This category is not catesian closed (if I am not mistaken that is because products do not commute with colimits in $\mathbf {Top}$, hence products cannot have right adjoints).
Nevertheless you can enrich $\mathbf{Top}$ over itself by regarding every set $\mathbf{Top}(A,B)$ as a topological space with the trivial topology (i.e. the topology in which every subset is open). With this topology the compositions are clearly continuous and it is trivial to verify that this is indeed a $\mathbf{Top}$-enriched category.
So while it is true that $\mathbf{Top}$ can be enriched over itself it is not cartesian closed.
Edit: from a discussion with the OP I have noticed that there is a misunderstanding on the notion of monoidal closed category.
Being monoidal closed is not equivalent to just having an internal $\hom$-functor, you also need for the covariant functors $\hom[X,-]$ to be right adjoint to the products $-\otimes X$.
The example above shows this: the products $-\times X$ in $\mathbf{Top}$ do not have, generally, right adjoints, but the $\hom$-functor can be turned into a $\mathbf{Top}$-valued functor.