Often we have adjoint pairs (A,B) (meaning A is left adjoint to B). Sometimes we have adjoint triplets (A,B,C) (meaning A is left adjoint to B and B is left adjoint to C. No adjoint relation between A and C obviously, since they have the same source and target).

So the first question is: can we have quadruplets (A,B,C,D) ? (meaning that C is also left adjoint to D).

Second question: It would be then possible that we have (A,D) besides (A,B). But I think not, because I think that it is not possible to have a functor A with two different right adjoints, B and D. Is this correct

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See this MO thread for some good examples for the first question. For the second question: prove that the right adjoint of a functor is unique up to a unique isomorphism. – t.b. Dec 16 '11 at 14:51

Actually, for every natural number $\ge 2$, there is a maximal string of adjoint functors of that length. Here's one construction: let $\mathbf{n}$ denote the poset category $\{ 0 < \ldots < n - 1 \}$. I claim there are functors (i.e. order-preserving maps) \begin{align} d_i & : \mathbf{n} \to \mathbf{n} + 1 & & 0 \le i \le n \newline s_i & : \mathbf{n} + 1 \to \mathbf{n} & & 0 \le i \le n - 1 \end{align} such that $$d_n \dashv s_{n-1} \dashv d_{n-1} \dashv \cdots \dashv s_0 \dashv d_0$$ is a maximal string of adjoint functors. Moreover, if $\mathbf{C}$ is a category with a terminal object but no initial object, there are functors \begin{align} \partial_i & : [\mathbf{n} + 1, \mathbf{C}] \to [\mathbf{n}, \mathbf{C}] & & 0 \le i \le n \newline \sigma_i & : [\mathbf{n}, \mathbf{C}] \to [\mathbf{n} + 1, \mathbf{C}] & & 0 \le i \le n \end{align} such that $$\partial_0 \dashv \sigma_0 \dashv \cdots \dashv \partial_n \dashv \sigma_n$$ is a maximal string of adjoint functors. (Here $[\mathbf{D}, \mathbf{C}]$ denotes the category of all functors $\mathbf{D} \to \mathbf{C}$.)
It is also not hard to find an infinite string of adjoint functors: for example, $$\cdots \dashv \textrm{id} \dashv \textrm{id} \dashv \textrm{id} \dashv \cdots$$ is such a string, though a little degenerate.
As for your second question, recall that adjoints are unique up to unique natural isomorphism, i.e. if $F \dashv G$ and $F \dashv G'$, then there is a unique natural isomorphism $G \cong G'$ which interacts nicely with the unit and counit of the adjunction.
There's something odd with the first example: you define $s_i$ with $0\leq i \leq n-1$ but then in the chain below you use $s_n$. Also, wouldn't such a chain prove only that there is a maximal string of adjoint functors of odd length? (As it stands, the chain has length $2n+1$) – lentic catachresis May 21 '12 at 14:17
@Bruno: $\sigma_n$ is not defined using $s_n$ (since it doesn't exist) and is quite different from the rest of them. If $\mathbf{C}$ has an initial object then there's also a $\sigma_{-1}$. Unless you are deliberately excluding posets from your definition of "category", the very first string of adjoints is a string of even length. – Zhen Lin May 21 '12 at 17:58
I never said anything about $\sigma_n$, I'm only talking about the first example (before the "moreover"): the second functor in the chain is $s_n$. – lentic catachresis May 22 '12 at 1:50
@Bruno: Sorry, that's a typo. It's $s_{n-1}$, as the pattern suggests. – Zhen Lin May 22 '12 at 7:21