Enriched functor categories and the embedding $\mathcal{B} \to \mathcal{B}^I$ Let $\mathcal{A}$ be an enriched category, tensored and cotensored over a closed symmetric (co)complete monoidal category $\mathcal{V}.$
Let $I$ be a small enriched category, and consider the category $\mathcal{A}^I$ of $\mathcal{V}$-functors and $\mathcal{V}$-natural tranformations; this category can be endowed with a $\mathcal{V}$-category structure where the enriched Hom object is defined as the enriched end $\underline{\int_{I}}\mathcal{A}(F(i),F(j)).$
$\mathbf{Question}$ 1

Is the the underlying set of $\underline{\int_{I}}\mathcal{A}(F(i),F(j))$ the set of $\mathcal{V}$-natural transformations?

This is equivalent to saying that the underlying category $(\mathcal{A}^I)_0$ of the enriched category $\mathcal{A}^I$ is the category of $\mathcal{V}$-functors and $\mathcal{V}$-natural tranformations; I am confident this is true (by analogy with the fact that in the $\text{Set}$ case the set of natural transformations can in fact be expressed as that end) but I don't have seen it written down in the usual references about enriched categories.
Question 2

If $I=\mathcal{C}[\mathcal{V}]$ is the free $\mathcal{V}$-category on
a small category $\mathcal{C},$ then the underlying set of
$\int_{I}\mathcal{A}(F(i),F(j))$ is just the set of natural
transformations between the associated functors $F,G: \mathcal{C} \to
 \mathcal{A}$ ?

Question 3
For unenriched categories, there always is a functor $\mathcal{B} \to \mathcal{B}^{\mathcal{C}}$ which takes $B \in \mathcal{B}$ to the constant functor at $B.$ Now suppose that $\mathcal{B}$ admits a structure of tensored and cotensored $\mathcal{V}$-category and that we put on $\mathcal{B}^{\mathcal{C}}$ the $\mathcal{V}$-category structure as above;

Does the functor $\mathcal{B} \to \mathcal{B}^I$ become enriched?


To state the question perhaps more precisely, does there exist a $\mathcal{V}$-functor $\mathcal{B} \to \mathcal{B}^I$ whose image by the forgetful functor $( \ )_0:\mathcal{V} \text{-CAT} \to \text{CAT}$ is the functor $B \mapsto (i\mapsto B )$ ?

In case the answer to questions 1,2 or 3 is negative I would like to know under which additional conditions on $\mathcal{V}$ it becomes true.
References to proofs in books/literature are very much appreciated.
 A: I follow Kelly [Basic concepts of enriched category theory].
Q1. In general, no; in practice, yes.
By definition, we have an equaliser diagram
$$\int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
and applying the underlying set functor preserves limits, so we get an equaliser diagram of sets:
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A}_0 (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
This is almost – but not quite – the definition of $\int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i))$.
For the latter we would have $\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j)))$ instead.
For every $V$ and $W$ in $\mathcal{V}$, the action of the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is a map
$$\mathcal{V}_0 (V, W) \longrightarrow \textbf{Set} (V_0, W_0)$$
so we get a canonical map:
$$\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j))) \longrightarrow \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j))) \tag{†}$$
This induces a comparison map
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \longrightarrow \int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i)) \tag{‡}$$
and it is easy to see that (‡) is a bijection if (†) is an injection.
This happens if the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is faithful, which is usually the case in practice.
(The main exception is when $\mathcal{V}$ is a category of multisorted structures, e.g. simplicial sets or chain complexes.)
Q2.
Yes.
When $V$ is a coproduct of $X$ copies of the monoidal unit, then the underlying set functor gives a bijection $\mathcal{V}_0 (V, W) \to \textbf{Set} (X, W)$.
Thus, if $\mathcal{I}$ is the free $\mathcal{V}$-category on an ordinary category, then (†) is a bijection, so (‡) is also a bijection.
(But we could have deduced this directly from the universal property of free $\mathcal{V}$-categories!)
Q3.
Yes.
Suppose $F$ and $G$ are constant functors, with values $A$ and $B$ respectively.
You can directly verify that the diagonal morphism $\mathcal{A} (A, B) \to \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i))$ equalises $\prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$ and so we have the desired morphism $\mathcal{A} (A, B) \to \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i))$.
(Again, you could instead use the universal property of free $\mathcal{V}$-categories here.
The point is that a $\mathcal{V}$-functor $\mathcal{A} \to [\mathcal{I}, \mathcal{A}]$ is essentially the same thing as a $\mathcal{V}$-functor $\mathcal{I} \to [\mathcal{A}, \mathcal{A}]$, and when $\mathcal{I}$ is free we can just take the constant functor with value $\textrm{id}_\mathcal{A}$.)
