Category with few endofunctors?

Question

Let $\mathcal C$ be a category whose skeleton has $\lambda$-many objects and $\kappa$-many morphisms. Then the skeleton of the endofunctor category $\mathcal C^{\mathcal C}$ has at most $\kappa^{3 \times \kappa}$ many morphisms.

My guess is that in most cases, this upper bound is achieved.

Question: What is an example of a category $\mathcal C$ (other than the terminal category or equivalents) whose skeleton has $\lambda$ many objects and $\kappa$ many morphisms, but the skeleton of whose endofunctor category $\mathcal C^{\mathcal C}$ has

1. Strictly fewer than $\kappa^\kappa$ many morphisms?

2. Strictly fewer than $\kappa^\kappa$ many objects?

3. As few as $\kappa$-many morphisms?

4. As few as $\lambda$-many objects?

EDIT: Neil Strickland's example of $B\mathbb N$ in the comments below affirmatively answers (1), (2), and (3). So it remains to see about (4).

Note that $\mathcal C^{\mathcal C}$ always has (skeletally) at least $\lambda+1$ many objects and $\kappa+1$-many morphisms, given by constant morphisms between constant endofunctors, along with the identity functor.

Note also that if $\mathcal C$ is accessible, and has as many objects and morphisms as the size of the universe, then the number of accessible endofunctors and morphisms between them is also the size of the universe. So if my guess is correct, then most endofunctors are usually non accessible.

See also here for an argument which shows that if $\mathcal C$ has small products and coproducts and is not a preorder, then $\mathcal C^{\mathcal C}$ has (skeletally) at least $2^\kappa$-many objects, where $\kappa$ is the size of the universe.

Answer 1

An example of (1),(2),(3),(4) with $\kappa=\lambda=|\mathbb R|$ is to take $\mathcal C$ to be the posetal category $\mathbb R.$ This is already skeletal. Its category of endofunctors is the poset of non-decreasing functions $f:\mathbb R\to\mathbb R,$ with the pointwise order. There are at most $|\mathbb R^{\mathbb Q}|=|\mathbb R|$ such functions: $f$ is determined by the preimages $f^{-1}((-\infty,q))$ for $q\in\mathbb Q,$ each of these intervals is convex, and there are at most $|\mathbb R|$ convex subsets of $\mathbb R.$

(I don't know if there are, unconditionally, examples with $\lambda$ and $\kappa$ regular. I think the $\mathbb R$ example generalizes to $2^\mu$ with the lex order, where $\mu$ is an ordinal satisfying $|2^{2^{<\mu}}|=|2^\mu|,$ for example $\mu=\beth_\alpha$ with $\alpha$ a limit ordinal. Here $2^{<\mu}$ means the functions $\mu\to 2$ with bounded support.)