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 $(3 \times \kappa)^\kappa$ 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.