A silly counterexample can be obtained by considering $\mathcal{V} = \mathrm{FinSet}_{\mathrm{bij}}$, the category of finite sets and bijections, with product of sets as the monoidal operation. Formally, the only enriched categories for this choice of $\mathcal{V}$ are trivial groupoids: a collection of objects where every hom-object is a singleton set $1$. This is because we require an enriched category to be equipped with 'identity morphisms'$1 \to \mathbb{C}(c,c)$ (which forces the latter to be isomorphic to $1$) and composition morphisms $\mathbb{C}(c',c) \otimes \mathbb{C}(c,c') \to \mathbb{C}(c,c)$, which forces each of the hom-objects in the domain to be isomorphic to $1$ also.
These are all equivalent, but as regards my question, the free symmetric monoidal $\mathcal{V}$-category on $\mathbb{C}$ exists: it is the trivial groupoid whose objects form the free monoid on the collection of objects of $\mathbb{C}$.
I say this example is 'silly' because the axioms of enriched categories mean that we're "actually" enriching over the trivial $1$-object category here, and if I enriched over that instead then my original observation becomes valid again (we just have $0=1$ in this case). To eliminate this case, let's generalize the above argument about which objects of $\mathcal{V}$ can actually arise as hom-objects.
Definition: An object $Z$ in $\mathcal{V}$ is a monoid mediator if there exist monoids $M,N$ in $\mathcal{V}$ and an object $Y$ such that $Z$ can be equipped with a right-$M$-action $Z \otimes M \to Z$ and a left-$M$-action $N \otimes Z \to Z$ and dually for $Y$, and there exist morphisms $Y \otimes Z \to M$ and $Z \otimes Y \to N$ respectively coequalizing (not universally) the maps $Y \otimes N \otimes Z \rightrightarrows Y \otimes Z$ and $Z \otimes M \otimes Y \rightrightarrows Z \otimes Y$.
Every hom-object $\mathbb{C}(c,c')$ in a $\mathcal{V}$-category is a monoid mediator, with $M = \mathbb{C}(c,c)$, $N = \mathbb{C}(c',c')$ and $Y = \mathbb{C}(c',c)$, their monoid structures and actions given by identities and composition; conversely given any monoid mediator we can construct a $2$-object $\mathcal{V}$ category in which is appears as a hom-object.
With this definition, we can replace $\mathcal{V}$ with its full subcategory on the monoid mediators. Using the fact that $\mathcal{V}$ was symmetric monoidal, we can deduce that this is a monoidal subcategory (and this operation is idempotent, since all monoids are monoid mediators). Note that if $\mathcal{V}$ was monoidal closed, then every object was already a monoid mediator of quite a special form!
Conjecture: If $Z$ is a monoid mediator in $\mathcal{V}$ then there exists a monoidal $\mathcal{V}$-category $(\mathcal{E},\oplus,I)$ and an object $X$ such that $\mathcal{E}(I,X) \cong Z$.
I believe I can prove this in the case that $\mathcal{V}$ is cartesian monoidal, since in that case we can assume that $M = N$ in the definition of monoid mediator by replacing both with $M \times N$ and projecting out the redundant component or inserting a unit in each of the morphisms demanded. Then we can construct a monoidal category in which every endomorphism monoid is $M \times N$ by gluing together an $\mathbb{N}$-indexed number of copies of the two-object category alluded to above. But I do not know if the above conjecture is realistic in general.
If the above conjecture holds, we can deduce (using the notation from the original question) that if the free category $(\mathcal{F},\oplus,I)$ on $\mathbb{I}$ exists, then $\mathcal{F}(I,*)$ must be weakly initial in the subcategory of monoid mediators. It seems that the morphism should be unique, but without a $\mathcal{V}$-indexed functorial family of witnesses we cannot deduce that $\mathcal{F}(I,*)$ forms a cone over the identity...
Modified question:
- Is there a symmetric monoidal category $\mathcal{V}$ in which every object is a monoid mediator but some object $X$ fails to appear as a hom-object of the form $\mathcal{E}(I,X)$ in a monoidal $\mathcal{V}$-category $(\mathcal{E},\oplus,I)$?
- Can $\mathcal{F}(I,*)$ fail to be initial even when the conjecture holds?