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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?

zeb
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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 $n>3$).

Noam D. Elkies
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    Ah, of course. Even $\mathbb{Z}/3$ acting on itself is a counterexample! – zeb Mar 30 '21 at 23:34
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    Yes, that's the first example of an "$a^2 x + b$ group" (for $n$ an prime power congruent to $3 \bmod 4$) that is transitive on pairs $-$ even simply transitive on pairs $-$ but not quite 2-transitive. – Noam D. Elkies Mar 30 '21 at 23:38