Adjunctions between discrete categories Exercise 2.1.13 asks:

What can be said about adjunctions between discrete categories?

First of all, what is "adjunction between categories"? Adjunctions between functors were defined on p.41. Is "adjunctions between discrete categories" supposed to mean "adjunctions between functors between discrete categories"?
Secondly, even if we interpret "adjunctions between discrete categories" as I wrote above, what kind of answer is expected from me? I don't think there is any "positive" result. For example, let $Ob(A)=\{a,a',a''\}, Ob(B)=\{b,b'\}$. Consider the functor $F:a,a'\mapsto b; a''\mapsto b'$ and $G:b\mapsto a', b''\mapsto a$. There is no bijection $B(F(a''),b')\cong A(a'',G(b))$.
 A: Yes, that is what's meant. When people say "an adjunction", they usually mean "a pair of functors $(F,G)$ with a natural isomorphism such that blabla"
Secondly, your example is an example of something that isn't an adjunction. 
In fact, there is a positive result : to get that result you have to answer these : 
1) What is a functor between discrete categories ? 
2) You have an isomorphism $A(a,G(b)) \simeq B(F(a),b)$. What does this imply with $b=F(a)$, in a discrete category ? 
(this 2) is a shadow of the more general notion of the unit of an adjunction, and the counit)
A: Here is the upshot of the discussion with Max. (Corrections are welcome.)
First of all, since Leinster only defines adjunctions between functors, the question should ask about adjunctions between functors between discrete categories.
Main Claim. Let  $\mathscr A$ and $\mathscr B$ be discrete categories and let $F:\mathscr A\to\mathscr B$ and $G:\mathscr B\to\mathscr A$ be functors. Then $F \dashv G $ iff $$(F:Ob(\mathscr A)\to Ob(\mathscr B), G:Ob(\mathscr B)\to Ob(\mathscr A))$$ is a pair of mutually inverse bijections.
Suppose $F\dashv G$. Then $$\mathscr B(F(A),B)\cong\mathscr A(A,G(B))$$ for all objects $A$ of $\mathscr A$ and $B$ of $\mathscr B$. In particular, if $B=F(A)$, then the LHS of the above has exactly one element. Thus so does the RHS. That is, there is exactly one arrow $A\to GF(A)$ in $\mathscr A$. This implies that $A=GF(A)$. Since $A$ was arbitrary, this shows that $GF=1_A$. Similarly one can show that $FG=1_B$. That is, $F$ and $G$ are mutually inverse bijections.
Now suppose that $F$ and $G$ are mutually inverse bijections. Consider an arrow $F(A)\to B$ in $\mathscr B$. The existence of this arrow implies $F(A)=B$. Now $A=GF(A)=G(B)$. Thus there is (exactly one, namely identity) arrow $A\to G(B)$ in $\mathscr A$. And conversely, if there is an arrow $A\to G(B)$ in $\mathscr A$, then $A=G(B)$, and $B=FG(B)=F(A)$, so there is a (unique) arrow $F(A)\to B$ in $\mathscr B$. This proves that there is a bijection $\mathscr B(F(A),B)\cong\mathscr A(A,G(B))$.
