This is somewhere between a comment and an answer. But it is too long for a comment, so I put it here.

To me the natural thing to look at is the action $G\curvearrowright G/H$. $|G/H|$ captures the index, which you surely want to do, and the action should in some sense capture the position of $H$ inside $G$. The issue then would be to try to come up with an appropriate notion of equivalence of these two objects. Here are 3 possibilities:

So let $H\subset G$ and $\Lambda\subset \Gamma$ be two inclusions.

1.) We can say $H\subset G \simeq_1 \Lambda\subset\Gamma$ if there is an isomorphism of groups $\Phi:G\rightarrow\Gamma$ such that $\Phi(H)=\Lambda$. This is the strongest notion of equivalence and would certainly imply that the two actions on the cosets spaces are the "same". The issue is that it requires the groups involved to be isomorphic, which you certainly don't want.

2.) We can forget about the acting group per se and only consider the action on the coset space (in particular look at the orbit equivalence relation). We define this equivalence relation by saying $g_1H\simeq g_2H\Leftrightarrow \exists g\in G$ such that $gg_1H=g_2H$. We do a similar thing on $\Gamma/\Lambda$. Then we say $H\subset G \simeq_2 \Lambda\subset\Gamma$ if there exists a bijection $\Phi:G/H\rightarrow \Gamma/\Lambda$ that takes equivalence classes to equivalence classes.

3.) The last one is exactly what you suggest for subfactors.

Just a quick final few comments. $1\Rightarrow 2$ but I'm not sure if $2\Rightarrow 3$. The motivation for these comes from three notions of equivalence you can give to a measure preserving action of a discrete group. In this case (for the analogous three equivalences) we have $1\Rightarrow 2\Rightarrow3$ and none of the implication are reversible, in general. Much of Popa's deformation/rigidity program is in trying to reverse these arrows in certain cases. In the ergodic theory case, $1\Rightarrow 2$ is obvious and $2\not\Leftarrow 1$ is not to hard. $2\Rightarrow 3$ isn't to hard either but $3\not\Leftarrow 2$ was quite difficult.

This is somewhere between a comment and an answer. But it is too long for a comment, so I put it here.

To me the natural thing to look at is the action $G\curvearrowright G/H$. $|G/H|$ captures the index, which you surely want to do, and the action should in some sense capture the position of $H$ inside $G$. The issue then would be to try to come up with an appropriate notion of equivalence of these two objects. Here are 3 possibilities:

So let $H\subset G$ and $\Lambda\subset \Gamma$ be two inclusions.

1.) We can say $H\subset G \simeq_1 \Lambda\subset\Gamma$ if there is an isomorphism of groups $\Phi:G\rightarrow\Gamma$ such that $\Phi(H)=\Lambda$. This is the strongest notion of equivalence and would certainly imply that the two actions on the cosets spaces are the "same". The issue is that it requires the groups involved to be isomorphic, which you certainly don't want.

2.) We can forget about the acting group per se and only consider the action on the coset space (in particular look at the orbit equivalence relation). We define this equivalence relation by saying $g_1H\simeq g_2H\Leftrightarrow \exists g\in G$ such that $gg_1H=g_2H$. We do a similar thing on $\Gamma/\Lambda$. Then we say $H\subset G \simeq_2 \Lambda\subset\Gamma$ if there exists a bijection $\Phi:G/H\rightarrow \Gamma/\Lambda$ that takes equivalence classes to equivalence classes.

3.) The last one is exactly what you suggest for subfactors.

Just a quick final few comments. $1\Rightarrow 2$ but I'm not sure if $2\Rightarrow 3$. The motivation for these comes from three notions of equivalence you can give to a measure preserving action of a discrete group. In this case (for the analogous three equivalences) we have $1\Rightarrow 2\Rightarrow3$ and none of the implication are reversible, in general. Much of Popa's deformation/rigidity program is in trying to reverse these arrows in certain cases. In the ergodic theory case, $1\Rightarrow 2$ is obvious and $2\not\Rightarrow 1$ is not to hard. $2\Rightarrow 3$ isn't to hard either but $3\not\Rightarrow 2$ was quite difficult.