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 $\mathcal{Q} = \bigcap_i \mathcal{R}^{G_i}$)