Let $A=(V, E)$ be a finite simple (no loops or multiple edges) graph. Let $G(A)$ be the following nilpotent group of class 2 and exponent $p$ (an odd prime). $G(A)$ as a set is $span(V)+span(E)$ where $span(X)$ is the elementary abelian group of exponent $p$ generated by the set $X$, and the commutator bracket is given by 1) the subgroup $span(E)$ is the center, 2) if $a,b\in V$ then $[a,b]=e\in E$ if $e$ is the edge connecting $a, b$ in $A$ or $=1$ if there is no edge $(a,b)$ or $(b,a)$. It is clear that every automorphism of the graph $A$ induces an automorphism of the group $G(A)$.

*Question 1*: Can $Aut(G(A))$ be much bigger than $Aut(A)$?

**Update** Derek Holt in his comments below showed that $Aut(G(A))$ always contains an elementary abelian normal subgroup of order $p^{|V||E|}$. Here is a more concrete question.

*Question 2*. Is it true that $Aut(G(A))$ comtains an involution iff $Aut(A)$ contains an involution?

This is related to Automorphism groups of odd order.

The positive answer to the next question would imply the positive answer to Question 2.

*Question 3.* Let $Q$ be the subgoup of $Aut (G(A))$ consisting of all automorphisms induced by the automorphisms of $A$. Is it true that $Aut(G(A))$ is equal to a semidirect product of $P$ and $Q$?

much. When the graph is the discrete with $n$ vertices, the numbers are easily computed and the size of $\textit{Aut}(G(A))$ (namely $[n]_p!(p-1)^np^{C(n,2)}$) is larger than the size of $\textit{Aut}(A)$ (which is only $n!$). If you demand nilpotence class exactly two, do the same construction but add one edge. The numbers should be about the same. $\endgroup$6more comments