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 $\mathbb{R}^n$.

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?

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.

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.

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.

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