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.