Let $G,H$ be infinite simple undirected graphs with the property that for any graph $X$ we have $|\text{Hom}(X,G)| = |\text{Hom}(X,H)|$. Does this imply that $G$ is isomorphic to a subgraph of $H$, and vice versa?


(Note that in the finite case the condition above implies $G\cong H$, see [here](http://www.maths.qmul.ac.uk/~pjc/csgnotes/hom1.pdf).)



Infinite graphs with "similar" Hom-sets
Did you want both G to be a subgraph of H and H to be a subgraph of G? Then I think the answer is negative. For example if G is infinite and have no isolated vertices and H is G together with an isolated vertex. If X have no isolated vertices the Hom-sets are the same. The set of maps from the set of isolated vertices of X to G or H should be of the same size as the cardinality of the vertex sets of G and H should be the same.


Edit: And if G for example is the integers with edges $\{i,i+1\}$ then I don't think H can be a subgraph.


