Isomorphism of directed graphs Consider the following directed graphs:


One is obtained from the other by reversing the direction of all edges.
Are they isomorphic as directed graphs ?
On the one hand, I would answer: no because there is no pair of maps between the vertices and the edges respectively that preserves the adjacency relation.
On the other hand, if one forgets about the ``labels'' on say, the edges, then the graphs are the same (just exchange the label 'e' with label 'f').
 A: Your comment helps clarify the source of your confusion:

A graph morphism is a pair of maps between the respective set of vertices $p:V \to V$ and and between the respective set of edges $q:E \to E$. If I set $q(e) = f$, $q(f) = e$ and $q(l) = l$ then because of the adjacency relation, I have to set: $w = \text{initial vertex of } f = \text{initial vertex of } q(e) = p(\text{initial vertex of e})= p(v)$. So $p$ has to exchange $v$ and $w$. Now $q(l) = l$ so again by the adjacency relation one must have: $v = \text{initial of vertex of } l = \text{initial vertex of } q(l) = p(\text{initial vertex of } l) = p(v)$. Hence $p$ has to fix the vertex $v$. This is a contradiction.

You implicitly (and wrongly) assume that edge $f$ in the left graph is the same as edge $f$ in the right graph. The confusion is only due to name clash. Renaming should save you from this mistake.
If we named edges in the right graph $e'$ and $f'$, then instead of writing this statement (original):

$q(e) = f$, ... $w$ = initial vertex of $f$

we would have written this statement:

$q(e) = f'$, ... $w$ = initial vertex of $f$

