= 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 $D\dashv O\dashv I$.
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 category>
$$
\boxed{C\dashv D\dashv O\dashv I.}
$$
It 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 $I(1\amalg1)$ 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. \b[Neither end of this chain has a further adjoint.]
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