(I wanted to post this a comment on @Will Sawin's answer, but I don't have the rep yet. Also it turned out to be convenient to flesh it out a little longer than comment.)

Will's answer assumes that you know a fixed element of $G/H$ that you can use as your basepoint for the fundamental group, namely the identity coset $H$. If you don't know that, you can't recover $H$ exactly, you can only recover $H$ up to conjugacy. (The permutation actions of $G$ on $G/H$ and $G/H'$ are permutationally isomorphic iff $H$ and $H'$ are conjugate subgroups.)

For example, in the case of $PSL(2,\mathbb{Z})$, this means that subgroups-up-to-conjugacy of index $n$ are in bijective correspondence with pairs of permutations $(a,b)$ on $[n]$ such that $a^{2} = b^{3} = 1$ (picking some natural generators of $PSL(2,\mathbb{Z}) \cong \mathbb{Z}_2 * \mathbb{Z}_3$)

and $\langle a, b\rangle$ acts transitively on $[n]$, up to the relation $(a,b) \sim (a',b')$ if there is a permutation $\pi \in S_{n}$ such that $\pi a \pi^{-1} = a'$ and $\pi b \pi^{-1} = b'$ (permutational isomorphism).

Subgroups (not up to conjugacy, but on the nose) are classified by the same pairs, but where the $\pi$ that we are allowed to conjugate by must fix, say, $1 \in [n]$ (corresponding to the identity coset of $H$ in $G/H$).