What functor is adjoint to the tensor product of 2-vector spaces? This is a refinement of my (naive, poorly asked) question here.  The reference for my question is Baez and Crans, HDA6.
Background: category objects, etc.
Let $\mathcal V$ be a category.  A category object internal to $\mathcal V$ consists of the following data and properties:

*

*Objects $C_0,C_1 \in \mathcal V$ and morphisms $s,t: C_1 \to C_0$ and $i: C_0 \to C_1$.

*Such that $s\circ i = t\circ i = \text{id}_{C_0}$ and the pull-back $C_1 \underset{C_0}{\times} C_1 = C_1 \underset{\displaystyle ^{\searrow\!\!^{\scriptstyle s}} {C_0} ^{^{\scriptstyle t} \!\!\swarrow} }{\times} C_1 $ exists.

*A morphism $m: C_1 \underset{C_0}\times C_1 \to C_1$ such that $s\circ m = s_R$ and $t\circ m = t_L$, where $s_R: C_1 \underset{C_0}\times C_1 \to C_0$ is the "$s$" projection from the right factor, and similarly for $t_L$.

*And such that the obvious "associativity" square (two ways to get from $C_1 \underset{C_0}\times C_1 \underset{C_0}\times C_1$ to $C_1$) and "identity" triangles (three ways to get from $C_1 = C_1 \underset{C_0}\times C_0 = C_0 \underset{C_0}\times C_1$ to $C_1$) commute.

For example, a category object in $\mathcal V = {\rm SET}$ is a small category.
For this post, I will be interested in $\mathcal V = {\rm VECT}$, the category of vector spaces over your favorite field.  I will call your favorite field "$\mathbb R$".
A category object in ${\rm VECT}$ is a 2-vector space.
2-vector spaces are relatively mild things.  Indeed, it turns out that in $\rm VECT$ the data and properties of 1-2 above uniquely determine a map $m$ satisfying 3-4.
By the general yoga known as "commutativity of internalization", a 2-vector space is the same as a "vector space object in $\rm CAT$".  More precisely, let $\rm CAT$ be the category of small categories.  Then it makes sense to talk about "field objects" — like a category object, a field object consists of some objects, some maps, some pull-backs, and some more maps, and some commuting diagrams.  In particular, by thinking of $\mathbb R$ as a discrete category ($\mathbb R_0 = \mathbb R = \mathbb R_1$ and $s = t = i = {\rm id}$), it is in fact a field object.  Then with some more diagrams, we can talk about "vector space objects over $\mathbb R$" internal to $\rm CAT$, and it is straightforward to check that these are the same as 2-vector spaces.
Background: tensor products
I know of two natural approaches to define "tensor products":

*

*Define a notion of "bilinear map", such that the assignment $X,Y,Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$ is contravariant in the first two spots and covariant in the last.  Then set $X\otimes Y$ to be the object (if it exists) that represents the function $Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$.

*Define a notion of "internal hom", i.e. a (nice) functor $\underline{\rm Hom}: \mathcal V^{\rm op} \times \mathcal V \to \mathcal V$.  For each $X\in \mathcal V$, define the functor $-\otimes X$ by declaring that it is left adjoint to $\underline{\rm Hom}(X,-)$.

Approach 1 is the way that tensor products are introduced in grade school.  Approach 2 is I think more standard in the real world.  We can implement each in the case of 2-vector spaces:

*

*The trick for approach 1 is that the notion of "bilinear" depends on more than just the category.  So realize the category of 2-vector spaces as the category of vector spaces objects in $\rm CAT$.  Recall that a morphism of 2-vector spaces is simply a morphism of underlying categories so that some diagrams commute.  Then we can say the following.  Let $X,Y,Z$ be 2-vector spaces.  Then a morphism of underlying categories $X \times Y \to Z$ is bilinear if a bunch of diagrams commute (these diagrams refer to the vector-space-object structures of $X,Y,Z$, and are precisely the diagrams that you learned in grade school).  By reproducing the proofs from vector spaces internal to $\rm SET$, this in fact defines a functor $\otimes$.

*Given 2-vector spaces $X,Y$, there is a category whose objects are linear functors $X \to Y$ and whose morphisms are linear natural transformations of functors, and this category has a natural structure as a 2-vector space.  Moreover, the corresponding notion of $\underline{\rm Hom}$ is correctly functorial, and has an adjoint.  So this approach defines a functor $\otimes$.

However:

The two "tensor products" defined in 1-2 above do not agree.

Writing 2-vector spaces $X = (X_1 \rightrightarrows X_0)$ and $Y = (Y_1 \rightrightarrows Y_0)$ as category objects in $\rm VECT$, approach 1 gives $(X\otimes Y)_a = X_a \otimes Y_a$ for $a=0,1$, with the tensor products of the structure maps.  Approach 2 also has $(X\otimes Y)_0 = X_0 \otimes Y_0$, but $(X\otimes Y)_1 \cong $
$$ X_0 \otimes Y_0 \oplus \text{coker}\Bigl( \bigl(\ker(X_1 \overset s \to X_0) \otimes \ker(Y_1 \overset s \to Y_0)\bigr) \overset{t\otimes{\rm id} - {\rm id}\otimes t}\longrightarrow \bigl( X_0 \otimes \ker(Y_1 \overset s \to Y_0) \oplus \ker(X_1 \overset s \to X_0) \otimes Y_0\bigr) \Bigr)$$
The structure maps are: $s$ is the projection onto the first factor, and $t$ is the sum of the same projection and the map ${\rm id}\otimes t + t\otimes {\rm id}$ from the second factor (it is well-defined out of the cokernel).
(There is probably a way to simplify the above description.  The trick is that, as explained in HDA6, the category of 2-vector spaces is equivalent as a category to the category of 2-term chain complexes, and this is the natural "internal Hom" over there.)
Anyway, a dimension count shows that these two "tensor" constructions are inequivalent in general.
Hence:
Question
Does the tensor product of 2-vector spaces given in "approach 1" above — the tensor product defined as representing "bilinear functors" — under this tensor product, does the functor $\otimes X$ have a right adjoint?
 A: I'll denote your category of 2-vector spaces by 2Vect.  By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on Δ≤1 where Δ≤1 denotes the full subcategory of Δ on the objects [0] and [1].  Therefore, colimits in 2Vect are computed objectwise under this identification.  So the functor – ⊗ X certainly has a right adjoint Hom(X, –) (by the adjoint functor theorem for locally presentable categories).  This adjunction also respects the Vect enrichment.
To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing Δi for the 2-vector space (Δi)j = HomΔ≤1([j], [i]) • ℝ, we have hom2Vect(Δi, X) = Xi as vector spaces, where hom denotes the Vect-enriched Hom.  So
Hom(X, Y)0 = hom(Δ0, Hom(X, Y)) = hom(Δ0 ⊗ X, Y) = hom(X, Y)
since Δ0 happens to be the unit for this ⊗, but
Hom(X, Y)1 = hom(Δ1 ⊗ X, Y)
will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).
