Let $G$ and $H$ be permutation groups on the natural numbers such that the orbits of $G$ and $H$ are all finite. Suppose that for all $\pi \in Sym(\mathbb{N})$, there is some $N$ (depending on $\pi$) such that for all $n \ge N$, the ordered tuple $(\pi(1),\pi(2),\dots,\pi(n))$ has a larger orbit (by a fixed ratio) under $G$ than it has under $H$.


Can $G$ and $H$ be conjugate in $Sym(\mathbb{N})$?


Edit: Answer is 'yes' (see Jim Belk's comment below); indeed $G$ can be conjugate to proper subgroups of itself of finite index, which makes the size of tuple orbit property automatic.


But what if $G$ only has finitely many orbits of size $n$ for each $n \in \mathbb{N}$? This would at least ensure that $G$ cannot be conjugate to one of its own subgroups.


Edit 2: An example would need to have the following property:


There is a tuple $t$, such that for any tuple $u$ for which $G\_u$ is contained in $G\_t$, then the $G$-orbit of $u$ is larger than the $H$-orbit of $u$.


So for instance if we pick a tuple $u$ by saying 'choose a large number $K$, then choose from among the $K$-tuples with no repeats one with smallest possible $G$-orbit', then $G\_u$ would not be contained in $G\_t$ no matter how large $K$ is. I think this rules out examples where the tuple stabilisers of $G$ are totally ordered, for instance if $G$ is cyclic and all orbits have length a power of a fixed prime.



Distinguishing finite-orbit permutation groups by action on tuples
Here's a case where $G$ and $H$ can be conjugate. First some notation: given a sequence $\{k\_n\}$ of positive integers, let $[k\_1,k\_2,\ldots]$ denote the permutation


$$(1,\ldots,k\_1)(k\_1+1,\ldots,k\_1+k\_2)(k\_1+k\_2+1,\ldots,k\_1+k\_2+k\_3)\cdots$$


with cycles of size $k\_1,k\_2,k\_3\ldots$. For example, $[1,1,1,1,\ldots]$ denotes the identity, $[2,2,2,2,\ldots]$ denotes $(1,2)(3,4)(5,6)(7,8)\cdots$, and $[2,3,2,3\ldots]$ denotes $(1,2)(3,4,5)(6,7)(8,9,10)\cdots$.


Let
$$g = [1,2,\;\;1,2,4,\;\;1,2,4,8,\;\;\ldots],$$
let
$$h = [1,1,1,\;\;1,1,1,2,2,\;\;1,1,1,2,2,4,4,\;\;\ldots],$$
and let $G$ and $H$ be the cyclic subgroups generated by these elements. Since $g$ and $h$ have the same cycle structure, they are conjuagte in $Sym(\mathbb{N})$, so $G$ and $H$ are conjugate subgroups.
However, for sufficiently large $n$, the orbit of $(\pi(1),\pi(2),\ldots,\pi(n))$ under $G$ will be precisely twice the size of the orbit under $H$.


Of course, in this example $G$ and $H$ both have infinitely many orbits of size $2^k$ for every $k$, so this does not answer the more restrictive version of the question.


