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](https://mathoverflow.net/a/177783/34538)), 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 $\mathcal{Q} = \bigcap\_i \mathcal{R}^{G\_i}$)



Infinite amenable group subfactors
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.)


