Preliminaries: in the following, $G$ will always be a finitely generated group and $\mathrm{Schr}(G,H;S)$ will be the Schreier graph corresponding to the symmetric generating set $S^{\pm1}$. Schreier graphs will be unlabelled non-oriented graphs, with possibly self-loops and multiple edges.
Before going further, observe that if we look at labelled Schreier graphs, the graphs is transitive (by labelled automorphisms) if and only if the subgroup is normal; so this is not really interesting.

# Transitive Schreier graphs

This question was somehow solved in https://arxiv.org/pdf/1505.03433.pdf

**Proposition**
A Schreier graph $\mathrm{Schr}(G,H;S)$ is transitive if and only if H is length-isomorphic to all of its conjugate. That is:
for every $g\in G$ there exists a group isomorphisms $\alpha_G\colon H\to g^{-1}Hg$ that preserves length: $|\alpha_G(h)|_S=|h|_S$ for every $h\in H$.

Observe that $\alpha_g$ is defined on $H$ only, not on $G$.

The above characterisation might seem silly and just an easy rewriting of what transitivity means for a Schreier graph. In some sense it is, however it can still be used to obtain some informations. A few words about that below.

# Dependence to the generating set

It is well-known (and easy to prove) that if $H$ is a normal subgroup of $G$, then all of its Schreier graphs are transitive. The converse is true and is in fact witnessed by small symmetric generating sets:

**Proposition**
If $\mathrm{Schr}(G,H;S)$ is transitive for all generating sets of size no more than $\mathrm{rank}$, then $H$ is a normal subgroup of $G$.

The above result shows that for non-normal subgroups, transitivity of the Schreier graph depends highly on the generating set. This is easily seen on finite groups, where every subgroup $H$ will have a transitive Schreier graph for $S=G\setminus\{1\}$. (In fact this is true for a general $G$, with the caveat that $S$ won't be finite in general)

# A strengthening of normality

In view of the above, let us say that a subgroup $H$ of $G$ is *length-transitive* if there exists a generating set of size no more than $\mathrm{rank}(G)+1$ such that $\mathrm{Schr}(G,H;S)$ is transitive. One can show that the intersection of two length-transitive subgroups is itself length-transitive.
This naturally lead to

**Definition**
A non-trivial group $G$ is *strongly simple* if its only length transitive subgroups are $\{1\}$ and $G$ itself.

It follows from the definition that cyclic groups of prime order are strongly simple and that strongly simple groups are simple. These implications are stricts as demonstrated by

**Proposition**
For odd $n\geq 7$, the alternating group $A_n$ is simple but not strongly simple.

**Proposition**
Tarski monsters are strongly simple (in fact, their Schreier graphs are not even quasi-transitive, and this for every finite symmetric generating set).

a prioriextend to the edges. $\endgroup$4more comments