Let $X$ be a set (in my case it is infinite) and $G,H \subset {\rm Sym\,} X$ (the symmetric group that acts on $X$). I know that $G$ has an orbit of a certain size (lets say $n$) and $H$ does not have an orbit of size $n$. I want to conclude that $G$ and $H$ are non-isomorphic.
I am not entirely sure that this is indeed true. After all, the isomorphism between $G$ and $H$ is only as groups and does not need to preserve the action in any way. I am sure that it does not preserve orbits itself, but does it preserve orbit sizes?
Subgroups $G,H\subseteq\mathrm{Sym}(X)$ can be isomorphic and have different orbit sizes.
For instance, given any two infinite multisets $\{\lambda_1,\lambda_2,\cdots\}$ and $\{\mu_1,\mu_2,\cdots\}$ of naturals one can create corresponding permutations $\pi$ and $\sigma$ with those cycle types. If one chooses them so one multiset has a value $n$ the other doesn't have, and yet they both have infinite $\mathrm{lcm}$s, then $G=\langle \pi\rangle$ and $H=\langle\sigma\rangle$ will be isomorphic but one will have an orbit of size $n$ and the other won't.
More generally, given any two $G$-sets $X,Y$ of the same size but different orbit sizes, we can biject them and treat it as two different actions of $G$ on $X$, in which case we get isomorphic copies of $G$ within $\mathrm{Sym}(X)$ having different orbit sizes. When possible, one way to get $X$ and $Y$ is picking a proper subgroup $H$ with $[G:H]<|G|$, then setting $X=H\times G/H$ and $Y=G$, so that $G$ acts trivially on $H$ to make sense of $X$, and acts regularly on $Y$.
Of course if $G$ and $H$ are conjugate they will have the same multiset of orbit sizes.