On p.333 of Conceptual Mathematics 2nd ed. we are asked to compute the exponential object $A^{A}$ in the category of irreflexive graphs, where $A$ is the graph of a single arrow s -> t.
(For context there is a graph $D$ that is the graph of a single dot *.
The book suggests using the properties of exponentials and graphs. For example to find the number of dots in $A^{A}$, find the number of maps $D \rightarrow A^{A}$, which is equivalent to the number of maps $A \times D \rightarrow A$. This is the same as the number of maps from the dots of $A$ to $A$, of which there are 4.
Similarly we can find the number of arrows by counting the maps of $A \rightarrow A^{A}$ which is the same as counting the maps of $A \times A \rightarrow A$ which there are 4, which are..
(s, s) and (t, t) must be mapped to s and t respectively, but (s, t) and (t, s) can be mapped to either s or t, hence 4.
So I know there are 4 dots and 4 arrows.. but I don't know how to map the arrows onto the dots.
Context: In the category of graphs an object is two sets $X$ (the edges of the graph) and $P$ (the nodes of the graph) equipped with two arrows $s, t: X \rightarrow P$ which correspond to the source and target of the edge. An arrow between two graphs is two arrows $f: X \rightarrow X'$ and $g: P \rightarrow P'$ which preserves source and target:
$$ g \circ s = s' \circ f $$ $$ g \circ t = t' \circ f $$
To see how the source and target mappings work, what you really want to do is, for a given edge $f:A\to A^A$, take the transpose of the composite $ev\circ id_A\times (f\circ\mathsf{src})$, where $\mathsf{src}$ is the inclusion of $D$ into the source of $A$'s edge. The composite is patently a morphism $A\times D\to A$, and its transpose is therefore a vertex $s_f:D\to A^A$. The resulting function $f\mapsto s_f$ works out to be the source function for the exponential graph.
If instead you use $\mathsf{tgt}:D\to A$ (let's call it), the inclusion of $D$ into the target vertex of $A$, then you have an analogous $f\mapsto t_f:D\to A^A$ forming a target function for the exponential graph.
More concretely, given a graph morphism $f:A\times A\to A$, let $f_V$ be the vertex part of the morphism. All the abstract nonsense above says is that the source function is $f\mapsto f_V(-,s)$ and the target function is $f\mapsto f_V(-,t)$.