TITLE: Is every permutation group on $n$ letters the symmetry group of a set of $n$ points in some euclidean space?
QUESTION [6 upvotes]: This question is superficially similar to a previous question. 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?

REPLY [14 votes]: 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 $n>3$).