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 $Sym(\mathbb{N})$?

It would seem strange if this happened, but I can't think of an invariant that would definitively distinguish $G$ from $H$.

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.