Infinite amenable group subfactors Let amenable groups $\Gamma$ and $\Gamma'$. They act outerly of only one manner on the hyperfinite ${\rm II}_1$-factor $\mathcal{R}$.   
Question:  $(\mathcal{R} \subset \mathcal{R}  \rtimes \Gamma)  \simeq (\mathcal{R} \subset \mathcal{R}  \rtimes \Gamma') \, \Rightarrow \, \Gamma \simeq \Gamma'$ ?   
If $\Gamma$ is infinite, then it exists an outer action on $\mathcal{R}$ such that  $\mathcal{R}^{\Gamma}  = \mathbb{C}$ (see comments of this answer), so $(\mathcal{R}^{\Gamma} \subset \mathcal{R})$ can't be the dual of $(\mathcal{R} \subset \mathcal{R}  \rtimes \Gamma)$.
 Let the infinite amenable group $\Gamma = \bigcup_iG_i$ with $(G_i)_i$ an increasing sequence of finite groups.
Now, $(\mathcal{R}^{G_i} \subset \mathcal{R})$ is the dual of $(\mathcal{R} \subset \mathcal{R}  \rtimes G_i)$ and $(\mathcal{R} \subset \overline{\bigcup_i\mathcal{R}  \rtimes G_i}) \simeq (\mathcal{R} \subset \mathcal{R}  \rtimes \Gamma)$.   
Question: How to characterize the subfactor $(\mathcal{Q} \subset \mathcal{R})$ for being the dual of $(\mathcal{R} \subset \mathcal{R}  \rtimes \Gamma)$?
(If I'm not mistaken  $\bigcap_i \mathcal{R}^{G_i} = \mathcal{R}^{\Gamma}$, so we can't take $\mathcal{Q} = \bigcap_i \mathcal{R}^{G_i}$)
 A: The answer to the first question is yes by a result of Herman and Ocneanu (MR1055223) or of Enock and Nest (MR1387518). 
We call an inclusion of II$_1$-factors depth 2 if $M_0'\cap M_2$ is abelian and $M_0'\cap M_3$ is a factor. (In fact, we only need to say $M_0'\cap M_3$ is a factor, and this works for arbitrary inclusions of factors, but I'll stick to the II$_1$ case.) Now for a depth 2 inclusion, Herman and Ocneanu say that the group $\Gamma$ can be recovered from unitaries which normalize $M_0$, using an extension of Proposition 1.7 of Pimsner-Popa's paper (MR0860811). But you should also check out Enock and Nest's article -- it's really very nice.
Now for the second part. There is not yet a completely satisfactory notion of the dual subfactor of an infinite index subfactor. Given an infinite index inclusion of II$_1$-factors $M_0\subset M_1$, the basic construction $M_2$ is a II$_\infty$-factor. So in this sense, it doesn't make sense to call $M_1\subset M_2$ the dual subfactor. This implies, as you noted above, that you can't get a II$_1$-factor from a downward basic construction either.
(However, Herman and Ocneanu have a nice notion of discrete and compact inclusions which are dual to each other. I'd recommend looking in their article.)
