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](https://mathoverflow.net/questions/334266/if-mathcal-c-mathcal-c-is-equivalent-to-mathcal-c-is-mathcal-c-nece/334962#334962) 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.



Category with few endofunctors?
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.)


