For any set $X$ we let $[X]^2 = \big\{\{x,y\}: x\neq y \in X\big\}$.


Let $G=(V,E)$ be a simple, undirected graph. Its open neighborhood hypergraph $\mathcal{H}(G)$ has the same vertex set $V$ with a hyperedge for the open neighborhood of every vertex $v \in V$. (The open neighborhood of $v\in V$ is the set $N\_v = \{y\in V: \{x,y\}\in E\}$.) 


Given a non-empty set $V$, are there $E\_1\neq E\_2\subseteq [V]^2$ such that ${\cal H}(V,E\_1) = {\cal H}(V,E\_2)$? Can $E\_1, E\_2$ even be chosen such that the graphs $(V,E\_1), (V,E\_2)$ are not isomorphic?



Different graphs with the same open neighborhood hypergraph
The answer to the first question is positive. Consider two graphs on eight vertices each consisting of two disjoint 4-cycles: the first one's cycles are $abcd$ and $efgh$, the second's ones are $afch$ and $ebgd$.


The answer for the second question is also positive, even if we consider $\mathcal H(G)$ to be a multi-hypergraph (thus taking neighborhoods with multiplicities). Take any connected non-bipartite graph $G$ and construct two graphs from it: first one is the disjoint union of two copies of $G$, another one is the tensor product of $G$ with an edge. Both have the same neighborhood multi-hypergraph, but one is not connected and another one is.


If you need, say, only connected graphs you may also obtain such by augmenting both graphs by a vertex connected to all vertices.


