Are there simple, undirected graphs $G, H$ that are non-isomorphic, but there exist graph homomorphisms $f\_1: G\to H$ and $f\_2: H\to G$ which are *bijective* set-maps $V(G)\rightarrow V(H)$ and $V(H)\rightarrow V(G)$?


**Notes.** 


* By the argument in Tobias Fritz's comment below, $G, H$ have to be infinite.
* As suggested by a commenter, one should make it unambiguously clear that here, 'simple, undirected graph'='irreflexive symmetric binary relation on a set'.



Non-isomorphic graphs with bijective graph homomorphisms in both directions between them
As vertex set, take $V=V'\cup V''$, the disjoint union of two infinite sets.


For $G$, take all edges except those joining pairs of vertices from $V''$.


For $H$, add one extra edge, between a pair of vertices $u,v\in V''$.


Then $G\not\cong H$, since if two vertices of $G$ are adjacent, then at least one of them is adjacent to every vertex, but that is not true for the vertices $u,v$ of $H$.


The identity map on $V$ is a bijective homomorphism $G\to H$.


There is a bijective homomorphism $H\to G$ given by choosing arbitrary bijections $V'\cup\{u\}\to V'$ and $V''\setminus\{u\}\to V''$.


