Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$.
Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the *class
transposition* $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$
which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$
and which fixes everything else.

> <b>Question:</b> Let $G < {\rm Sym}(\mathbb{Z})$ be a group generated by
> $3$ class transpositions, and assume that the integers $0, \dots, 42$
> all lie in the same orbit under the action of $G$ on $\mathbb{Z}$.
> Is the action of $G$ on $\mathbb{N}_0$ necessarily transitive?

*Remarks:*

  - When replacing $42$ by $41$, the answer obviously gets negative since
    the finite group
    $$
      G \ := \ \langle \tau_{0(2),1(2)}, \tau_{0(3),2(3)}, \tau_{0(7),6(7)} \rangle
    $$
    acts transitively on the set $\{0, \dots, 41\}$.
    Therefore if true, the assertion is sharp.

  - There is computational evidence suggesting that there is, say,
    "a reasonable chance" that the answer is positive.

  - A positive answer would mean that groups generated by $3$ class
    transpositions are "well-behaved" in the sense that for deciding
    transitivity, looking at very small numbers is sufficient, and that
    for larger numbers "nothing can happen any more".

  - A positive answer would imply the [Collatz conjecture][2].
    On the other hand, if the Collatz conjecture holds, this would (by far!)
    not imply a positive answer to the question.

  - There is a related question [here][1].

  [1]: http://mathoverflow.net/questions/112527/groups-generated-by-3-involutions
  [2]: https://en.wikipedia.org/wiki/Collatz_conjecture