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.