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 $k_1,k_2,k_3ldots$. 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 $[2,3,2,3ldots]$ 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.
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