Let $G$ and $H$ be graphs on the vertex set $\{1, \ldots, n\}$ and let $(e\_i)$ be the standard basis of $\mathbb{R}^n$. For each edge $\{i,j\}$ define *edge vectors* $e\_i - e\_j$ and $e\_j - e\_i$ in $\mathbb{R}^n$.



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> Question 1: If there is a linear isomorphism of $\mathbb{R}^n$ with itself that takes the edge vectors of $G$ bijectively onto the edge vectors of $H$, must $G$ and $H$ be isomorphic?
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The answer to this question is no: if $G$ and $H$ are both trees then there is such a linear isomorphism. Aside from $e\_i - e\_j$ being the negation of $e\_j - e\_i$, there are no linear dependences among the edge vectors, and the edge vectors of $G$ can be mapped to the edge vectors of $H$ in any manner.



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> Question 2: Same as Question 1, but now assuming that $G$ and $H$ both have central vertices, i.e., each of them has a vertex which is adjacent to every other vertex.
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I assumed a counterexample to Question 1 would easily yield a counterexample to Question 2, but I don't see this. A counterexample to Question 2 is what I need.



Non-isomorphic graphs with isomorphic edge vectors
Attempt to show that there is no example for Question 2. 


Let $G$ be a graph with central vertex $v\_0$, $H$ be a graph with central vertex $u\_0$ and cycle structures (cyclic matroids) of $G$ and $H$ are isomorphic. I claim that $H$ and $G$ themselves are isomorphic as graphs. Let $T$ be a spanning tree in $G$ formed by edges incident to $v\_0$. It corresponds to some spanning tree $f(T)$ in $H$, here $f$ is an isomorphism of matroids (so, $f$ is defined on edges of $G$). Note that in $G$ any edge $e\notin T$ belongs to a triangle with two edges from $T$. Thus the same holds in $H$. Apply this to edges in $H$ incident to $u\_0$ but not coming from $u\_0$. We see that maximal path in $f(T)$ going from $u\_0$ consists at most two edges. Let $u\_0u\_1,\dots,u\_0u\_k$ be edges incident to $u\_0$ and belonging to $f(T)$, $v\_0v\_i$ be their $f$-preimages. Next, if $u\_i$, $1\leqslant i\leqslant k$, is incident to some edge $u\_iu\_m\in f(T)$, $m>k$, then denote by $v\_0v\_m$ $f$-preimage of the edge $u\_0u\_m\in H$. Then $v\_i\rightarrow u\_i,i=0,1,\dots$ is isomorphism of $G$ and $H$.


