Constant enriched functors need not exist

Let $\mathcal{I}$ be the one-object $\mathbf{A}\mathbf{b}$-category with hom-object $\mathbb{Z}$, so its unique object has endomorphism object the tensor unit. Let $\mathcal{T}$ be the one-object $\mathbf{A}\mathbf{b}$-category with hom-object $0$, the terminal object of $\mathbf{A}\mathbf{b}$; the identity map $\mathbb{Z}\to 0$ and the zero composition make this a valid $\mathbf{A}\mathbf{b}$-category by [L3].

The underlying ordinary category ${\mathcal{T}}_0$ has one object and one morphism, because $\mathbf{A}\mathbf{b}(\mathbb{Z},0)$ is a singleton by [L2]. The underlying ordinary category ${\mathcal{I}}_0$ also has one object, with morphism set $\mathbf{A}\mathbf{b}(\mathbb{Z},\mathbb{Z})$. Sending the unique object of ${\mathcal{T}}_0$ to the unique object of ${\mathcal{I}}_0$ and its identity to $1_{\mathbb{Z}}$ therefore defines an ordinary constant functor ${\mathcal{T}}_0\to {\mathcal{I}}_0$.

A $\mathbf{A}\mathbf{b}$-enriched functor $\mathcal{T}\to \mathcal{I}$ would need a hom-object map $0\to \mathbb{Z}$ preserving the enriched identity. But the identity in $\mathcal{T}$ is the unique map $\mathbb{Z}\to 0$, and the identity in $\mathcal{I}$ is $1_{\mathbb{Z}}:\mathbb{Z}\to \mathbb{Z}$; preserving identities would force the composite $\mathbb{Z}\to 0\to \mathbb{Z}$ to be $1_{\mathbb{Z}}$, impossible because the composite through $0$ is the zero homomorphism. This contradicts [L1].

Hence the ordinary constant functor of step 2.1 has no enriched lift, so constant enriched functors need not exist.