It has $2^{2^{\aleph_0}}=2^c$ pairwise non-isomorphic subgroups (this is the maximum we could expect since this is the number of subsets).

To prove this, first start with a family of groups $(G_i)_{i\in c}$ such that

 - all $G_i$ are countable, torsion-free, with trivial center, and directly indecomposable
 - the $G_i$ are pairwise non-isomorphic.

The existence of such a family (even with $G_i$ finitely generated) is standard (just start with $H_i$ torsion-free countable, and define $G_i=H_i\ast H_i$ which will satisfy in addition the other two conditions).

For $I\subset c$, define $G_I=\bigoplus_{i\in I}G_i$, where $\bigoplus$ denotes the restricted direct product.

The $G_I$ are pairwise non-isomorphic, because being center-free and indecomposable ensures the uniqueness of the restricted direct product decomposition.

However, $G_I$ does **not** embed into $\mathrm{Sym}(\omega)$, so this is not the example yet.

Embed $G_I$ into $\mathrm{Sym}(\omega)/\mathrm{Sym}_{\mathrm{fin}}(\omega)$. This is possible as follows: consider $c$ infinite countable subsets $(A_i)$ of $\omega$, with pairwise finite intersection (the classical trick is to identify $\omega$ with $\mathbf{Q}$ and choose for every real the image of a sequence converging to it). Embed $G_i$ into $\mathrm{Sym}(A_i)$; the intersection property implies that this extends to a homomorphism $f$ of $G=\bigoplus_{i\in c}G_i$ into $\mathrm{Sym}(\omega)/\mathrm{Sym}_{\mathrm{fin}}(\omega)$ (quotient by the group of finitely supported permutations). Then $f$ is injective (because each $G_i$ is torsion-free). Now, for each $I$, let $H_I$ be the inverse image of $f(G_I)$ in $\mathrm{Sym}(\omega)$. Then the set of torsion elements in $H_I$ is $\mathrm{Sym}_{\mathrm{fin}}(\omega)$ and the quotient is isomorphic to $G_I$; in particular, the $H_I$ are pairwise non-isomorphic when $I$ ranges over subsets of $c$.