A bijection on underlying hom-sets need not exhibit a cotensor

Give $E$ its strict monoidal structure induced by addition: it has one object, both composition and tensor of endomorphisms are addition in $\mathbb{N}$, and commutativity gives the interchange law. Hence there is a one-object $\mathbf{C}\mathbf{a}\mathbf{t}$-enriched category $\mathcal{B}$ with sole object $C$, hom-category $\mathcal{B}(C,C)=E$, and enriched composition given by this tensor.

The underlying category ${\mathcal{B}}_0$ has one object and one morphism, since the objects of $E$ form a singleton. Thus, for its only test object $B=C$, there is a bijection ${\mathcal{B}}_0(B,C)\cong {\mathbf{C}\mathbf{a}\mathbf{t}}_0(X,\mathcal{B}(B,C)):$ both sides are singletons, because a functor from the discrete two-object category $X$ to the one-object category $E$ is unique on objects and identities. This bijection is automatically natural in the one-object category ${\mathcal{B}}_0$.

If $C$ were its own cotensor by $X$, [L1] would require an isomorphism of categories $E=\mathcal{B}(C,C)\cong [X,E]\cong E\times E\mathrm{.}$ But the endomorphism monoids of the unique objects are respectively $\mathbb{N}$ and ${\mathbb{N}}^2$, which are not isomorphic: the former has one indecomposable nonzero generator and the latter has two. Hence the natural underlying hom-set bijection of step 2.1 does not exhibit a cotensor.