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.