This question is superficially similar to a [previous question](https://mathoverflow.net/q/993/2363). Suppose I am given a permutation group $G \subseteq S\_n$. Is it always possible to find a set $X$ of $n$ points in $\mathbb{R}^n$, such that the isometry group of $X$ (together with its natural action on $X$) is equal to $G$ *as a permutation group*?



Is every permutation group on $n$ letters the symmetry group of a set of $n$ points in some euclidean space?
No. If $G$ is $2$-transitive $-$ or even transitive on pairs $-$ then
$X$ must be the set of vertices of a regular simplex,
which has isometry group $S\_n$. But there are plenty of
examples of $2$-transitive groups $G$ properly contained in $S\_n$
(such as the $ax+b$ group if $n$ is a prime power and $n>3$).


