 Solution

Let $D(S)$ be the discrete category on $S$, and $I(S)$ the indiscrete category, with exactly one morphism between each ordered pair of objects. A functor $D(S)\to \mathcal{A}$ is precisely a map $S\to O(\mathcal{A})$; a functor $\mathcal{A}\to I(S)$ is precisely a map $O(\mathcal{A})\to S$. Thus $D⊣O⊣I$.

Let $C(\mathcal{A})={\pi }_0\mathcal{A}$ be the set of connected components, where objects are related by finite zigzags of morphisms in either direction. A functor $\mathcal{A}\to D(S)$ must take the endpoints of every morphism to the same point, so it is exactly a map ${\pi }_0\mathcal{A}\to S$. This gives the adjoint chain for the objects of a categoryIt cannot extend to the left. The two functors from the terminal category to $I(\{0,1\})$ selecting distinct objects have empty equalizer in $\mathbf{Cat}$. Applying $C$ gives two identical maps between singleton sets, whose equalizer is a singleton. Thus $C$ does not preserve equalizers and cannot be a right adjoint.

It cannot extend to the right either. The coproduct of two copies of $I(1)$ in $\mathbf{Cat}$ is the discrete two-object category. But $I(1⨿1)$ is the indiscrete two-object category, with morphisms between its distinct objects. They are not isomorphic, or even equivalent. Hence $I$ does not preserve binary coproducts and cannot be a left adjoint. Neither end of this chain has a further adjoint.