Small "transitive" subsets of transitive groups Let $G$ be a group and $\varphi$ be a transitive group action of $G$ on $\{1, 2, \dots, n\}$. Does a subset $A\subseteq G$ such that $|A|=n$ and $A$ "acts transitively" on $\{1, 2, \dots, n\}$ always exist? Since transitive group actions are usually defined only for groups what I mean by $A$ "acting transitively" is that $\forall i, j \in \{1, 2, \dots, n\}: \exists f \in A: \varphi(f,i) = j$.
I suspect that the answer is negative but I have not been able to find a counterexample so far.
 A: I think a not-terribly-easy counterexample can be found in this paper

Theo Grundhöfer and Peter Müller, Sharply 2-transitive sets of permutations and groups of affine projectivities. Beiträge Algebra Geom. 50 (2009), no. 1, 143–154.

In Theorem 1.7 one find the statement

Let $G$ be the Conway group Co3 in its doubly transitive action of
  degree 276. Then the stabilizer $G_\omega$ of degree 275 has no sharply transitive subset.

A sharply transitive subset should be exactly a set like the $A$ in the question.
A: $S_4$ has a transitive action of degree $6$, with point-stabiliser given by $\langle (1,2),(3,4)\rangle$. This has no regular subgroup (because the stabiliser does not have a complement, which is easy to see, since the only subgroups of order $6$ in $S_4$ are $S_3$'s, and these all intersect the point-stabiliser non-trivially).
The index $2$ subgroup $A_4$ is also transitive, and so also does not have a regular subgroup. These are the two smallest examples of transitive groups with no regular subgroups.
I just did a computer check, and they also don't have regular subsets. 
Here's a short human-checkable proof of this fact in the case of $A_4$:
This action of $A_4$ on $6$ points is imprimitive, with three blocks of size $2$. The double transpositions preserve the blocks, fixing one point-wise and swapping the elements in the other two, the elements of order three permute the blocks.
We are looking for a regular subset $X$, which must have size $6$. 
The eight $3$-cycles have only four distinct images for a particular point (since they always map a point outside its block), so there can't be more than four $3$-cycles in $X$.
This means that $X$ contains at least two non-$3$-cycles, but all these have images in common (the double transposition fix some points, so agree with the identity on those, and each pair of double transposition has a block in common which they swap), contradiction.
