Is the invariant subalgebra of the von Neumann algebra $L(F_k)$ isomorphic to $L(F_k)$? Let the symmetric group $G=S_{k}$ act on the von Neumann algebra of the free group $L(F_k)$ via permuting its generators. Is the fixed point algebra under the action isomorphic to the whole algebra, i.e., $L(F_k)^{S_k}\cong L(F_{k})$?
 A: No, this is not the result that you should expect. Below, I prove that $L(F_k)^{S_k} \cong L(F_N)$ where $N = 1 + (k-1) |S_k| = 1 + (k-1) \cdot k!$. Of course, we do not know if $L(F_k) \not\cong L(F_N)$, but that is independent of the question.
In analogy with the Nielsen-Schreier index formula, one expects that a subfactor of index $m$ in $L(F_k)$ is isomorphic with $L(F_{1+(k-1)m})$, but proving this is an open problem. Note that this should even hold for non-integer indices $m$ by using interpolated free group factors.
For the specific example in the question, this can however be worked out. Slightly more generally, one may consider a transitive and faithful action of a finite group $\Gamma$ on a finite set $I$ and prove that $L(F_I)^\Gamma \cong L(F_N)$ where $N = 1 + (|I|-1)|\Gamma|$. To prove this formula, denote $A = L(\mathbb{Z})$ equipped with its natural trace. View $L(F_I)$ as the free product of $|I|$ copies of $A$. Fix an element $i_1 \in I$ and denote by $\Gamma_1 < \Gamma$ the stabilizer of $i_1$. The crossed product $L(F_I) \rtimes \Gamma$ is then naturally isomorphic with the amalgamated free product
$$M = (A \otimes L(\Gamma_1)) *_{L(\Gamma_1)} L(\Gamma) \; .$$
Since $\Gamma \curvearrowright I$ is faithful, the action $\Gamma \curvearrowright L(F_I)$ is outer and $M$ is a $II_1$ factor. By Theorem 4.2 of [DK11], $M$ is isomorphic with an interpolated free group factor $L(F_s)$ and, using the notation of [DK11], the value of $s$ can be computed as
$$s = \operatorname{fdim}(A \otimes L(\Gamma_1)) + \operatorname{fdim}(L(\Gamma)) - \operatorname{fdim}(L(\Gamma_1)) \; .$$
Since $A \otimes L(\Gamma_1)$ is diffuse, one has $\operatorname{fdim}(A \otimes L(\Gamma_1))=1$. For any finite group $\Lambda$ and using the canonical trace on $L(\Lambda)$, one has that
$$\operatorname{fdim}(L(\Lambda)) = 1 - \frac{1}{|\Lambda|} \; .$$
We thus conclude that
$$s = 1 + \frac{1}{|\Gamma_1|} - \frac{1}{|\Gamma|} \; .$$
So we have proven that
$$L(F_s) \cong M \cong L(F_I) \rtimes \Gamma \cong M_{|\Gamma|}(\mathbb{C}) \otimes L(F_I)^\Gamma \; .$$
We conclude that $L(F_I)^\Gamma \cong L(F_s)^{1/|\Gamma|} \cong L(F_N)$ where
$$N = 1 + |\Gamma|^2 (s-1) = 1 + (|I|-1) |\Gamma| \; .$$
[DR11] K.J. Dykema and D. Redelmeier, The amalgamated free product of hyperfinite von Neumann algebras over finite dimensional subalgebras. Houston J. Math. 39 (2013), 1313–1331.
