There are a couple of cases that we can dispose of straight away.
A self-edge on some vertex $v$ is an edge $v \rightarrow v$. If $H$ has even one vertex with a self-edge, then it's trivial to construct a homomorphism which maps all vertices to that vertex and all edges to that edge.
Similarly, if $H$ doesn't have a single vertex with a self-edge but $G$ has, no homomorphism exists.
Finally, we can deal with the case where $G$ has no self-edges. Then $G$ must be a chromatic graph: for some $k$, the vertices of $G$ can be assigned $k$ colours such that the start and end of each edge do not share the same colour.
Denote the complete graph with $k$ vertices by $K_k$.
Exercise: Show that a homomorphism $f : G \rightarrow K_k$ describes a $k$-colouring of $G$. Conversely, show that every $k$-colouring of $G$ can be represented as a homomorphism $f : G \rightarrow K_k$.
So any solution to your problem is also a solution to the $k$-colouring problem.
The decision problem of testing a mapping to see if it is a graph homomorphism is trivially in P. It follows that your problem is FNP-complete.