Fundamental group and group measure space construction Let $N$ be a type ${\rm II}$ factor, with trace $\tau$. Consider its fundamental group$$ \mathcal{F}(N)= \{ \tau(p)/\tau(q) \ | \ p,q \text{ non-zero finite projections in } N \text{ and } pNp \simeq qNq  \}. $$
Let $\alpha$ be a free ergodic measure preserving action of a countable ICC group $\Gamma$ on a $\sigma$-finite standard Borel measure space $(X,\mu)$.  Then  $L^{\infty}(X,\mu) \rtimes_{\alpha} \Gamma$  and  $L(\Gamma)$ are type ${\rm II}$ factors.    

Question:  Is $\mathcal{F}(L^{\infty}(X,\mu) \rtimes_{\alpha} \Gamma)$ a subgroup of $\mathcal{F}(L(\Gamma))$?

I did not find a counterexample in the following reference: On the fundamental group of ${\rm II}_1$ factors and equivalence relations arising from group actions, by Sorin Popa and Stefaan Vaes.  
Application: A positive answer would solve the free group factor isomorphism problem.
Proof: The group measure space construction $\mathcal{M} = L^{\infty}(\mathbb{S}^{1}, Leb) \rtimes_{\alpha} \mathbb{F}_{2}$ in this post is a ${\rm III}_1$ factor, so that its core $\widetilde{\mathcal{M}} = \mathcal{M} \rtimes_{\sigma} \mathbb{R} = L^{\infty}(\mathbb{S}^1 \times \mathbb{R}_{+}^*, Leb) \rtimes_{\widetilde{\alpha}} \mathbb{F}_2$ (see this answer) is a ${\rm II}_{\infty}$ factor of fundamental group $\mathbb{R}_{+}^*$ (moreover $\widetilde{\alpha}$ is free and ergodic). But by assumption $\mathcal{F}(\widetilde{\mathcal{M}})$ would be a subgroup of $\mathcal{F}(L(\mathbb{F}_{2}))$, so that $\mathcal{F}(L(\mathbb{F}_{2})) = \mathbb{R}_{+}^*$ also, implying that for all $n \ge 2$, $L(\mathbb{F}_{n}) \simeq L(\mathbb{F}_{2})$, by the works of Voiculescu and Radulescu.
Bonus question: Can every subgroup of $\mathcal{F}(L(\Gamma))$ be realized as $\mathcal{F}(L^{\infty}(X,\mu) \rtimes_{\alpha} \Gamma)$?
For this bonus question, I expect at most a counter-example (because a proof could be very hard).  

Naive approach for a positive answer to the main question:  
Let $N$ be $L^{\infty}(X,\mu) \rtimes_{\alpha} \Gamma$, as specified above. First of all, $L(\Gamma)$ can be taken as a subfactor of $N$. Take $t \in \mathcal{F}(N)$, then there are projections $p,q \in L(\Gamma) \subset N$ such that $\tau(p)/\tau(q) = t$. Then, by definition of the fundamental group, $pNp$ is isomorphic to $qNq$ (because the isom. class of such compression depends only on the trace of the projection). Let $\Phi: pNp \to qNq$ be an isomorphism. Then $\Phi(pL(\Gamma)p)$ = $qKq$, for some $K \subset N$.
Can we choose $\Phi$ such that we can take $K = L(\Gamma)$?
If so, $t$ is in $\mathcal{F}(L(\Gamma))$, and the result follows.
 A: I asked Stefaan Vaes by email, below is his answer (reproduced with his authorization):  

For instance, take $G$ to be the semidirect product of $\mathbb{Z}^2$
  and ${\rm SL}(2,\mathbb{Z})$. The group von Neumann algebra $L(G)$ has
  trivial fundamental group and this was even the very first ${\rm
 II}_1$ factor that was known to have trivial fundamental group (paper
  of Popa in Annals of Math). Now fix a prime number $p$ and take the
  essentially free, ergodic, pmp action of $G$ on $\mathbb{Z}_p^2$
  (where $\mathbb{Z}_p$ are the $p$-adic integers). The action of $G$ on
  $X=\mathbb{Z}_p^2$ is given as follows: $\mathbb{Z}^2$ acts by
  translation and ${\rm SL}(2,\mathbb{Z})$ acts in the "obvious" way.
Then consider the subset $Y$ of $X$ given by $Y = (p \mathbb{Z}_p)^2$.
  The restriction of the orbit equivalence relation to $Y$ is precisely
  the orbit equivalence relation of the semidirect product of $(p
 \mathbb{Z})^2$ and ${\rm SL}(2,\mathbb{Z})$. So, multiplication by $p$
  will make sure that $p^2$ belongs to the fundamental group of the
  orbit equivalence relation, and thus to the fundamental group of the
  crossed product ${\rm II}_1$ factor.
Conclusion: the crossed product has a nontrivial fundamental group,
  but $L(G)$ has trivial fundamental group.

Now, the topology of $\mathbb{Z}_p$ is that of a Cantor set. Then:  
Can we still hope that the question has a positive answer if we restrict to a more "regular" case like a continuous action on a topological manifold (with or without boundary), or even (if necessary) a smooth action on a smooth manifold (with or without boundary)?  
Here is an other answer of Stefaan:  

When it comes to von Neumann algebras or ergodic equivalence
  relations, there is no way to "see" such topological or geometric
  properties of group actions. So I am sure that there are also
  counterexamples that have natural, nice topological or geometric
  models. It is just a bit harder to find such counterexamples.

