This is another Tom Leinster problem.
Let $O:Cat \rightarrow Set$ be the functor taking a small category to its set of objects. Exhibit a chain of adjoints $$C \dashv D \dashv O \dashv I.$$
Well, for $I,$ we want a natural isomorphism $Hom(O(C),S) \cong Hom(C,I(S)).$ Let $I:Set \rightarrow Cat$ send any set $S$ to the category with objects the elements of $S$ and exactly one morphism $f:s \rightarrow s'$ for any $s,$ $s' \in S.$ This means that a functor $F:C \rightarrow I(S)$ for any category $C$ is determined entirely by its behavior on objects, and so there's a correspondence between $Hom(O(C),S)$ and $Hom(C,I(S)).$ Naturality is easy to check.
By similar reasoning, $D:Set \rightarrow Cat$ sends a set $S$ to the discrete category with objects $d \in S,$ and $C:Cat \rightarrow Set$ sends a category $C$ to its set of connected components.
If I'm wrong about any of those, I would welcome corrections, but I'm mainly stating them as background to the next part of the question, which is what I'm struggling with.
The question is whether this chain of adjoints can be extended further in either direction. The hint provided is to use the fact that left adjoints preserve terminal objects, and right adjoints preserve initial objects.
My first thought was to assume that $I$ has a right adjoint for contradiction. Then $I$ must send the terminal object of $Set$ to the terminal object of $Cat.$ But $I$ applied to the one element set yields the one element category with a single morphism, which is the terminal object of $Cat.$ Similarly, $C$ sends the initial object of $Cat$ to the initial object of $Set.$
I was assuming that the answer to the question would be that the chain couldn't be extended, since I have no idea how to use the hint to prove the existence of further adjoints. Am I perhaps misunderstanding something?