A regular subset of a transitive group

Question

Given a group $G$ acts transitively on a finite set $X$, we know that it is not necessarily true that it has a transitive subgroup of order $|X| $(transitive subgroup of an action). However, I was wondering if $G$ has a subset $S$, with order $|X|$, such that for all $x$, $y$ in $X$, there is a unique element of $S$ that maps from $x$ to $y$?

Answer 1

Expanding on a comment above by David A. Craven.

We will show a counter-example.

The elements of $A_4$ are:

$$e,\\(123),(132),\\(124),(142),\\(134),(143),\\(234),(243),\\(12)(34),(13)(24),(14)(23).$$

They act on $X,$ the $2$-subsets of $\{1,2,3,4\},$ inherited by their permutations. The six $2$-subsets are:

$$12,13,14,23,24,34$$

Any transitive subset of $A_4$ must contain one of the elements of order $2,$ since we need $12\mapsto 34.$ Only $(13)(24)$ or $(14)(23)$ do that, and we can only have one of them if we require uniqueness.

But if our transitive subset has the uniqueness property, then $e$ cannot be in the transitive subset, since $13$ is fixed by $(13)(24)$ and $14$ is fixed by $(14)(23).$

But the elements of order $3$ do not fix any of the elements of $X.$ So we need each of the elements of order $2$ to have permutations in our transitive set which send $x\mapsto x$ for each $x\in X.$

But we already saw that we can’t have two elements of order $2$ if we require uniqueness.

So there is no such set.