Let $D$ be an infinite UHF algebra, e.g. the infinite tensor product of the matrix algebra $M_k(\mathbb{C})$. The permutation group $\Sigma_n$ acts on the $n$-fold tensor product $D^{\otimes n}$ in a canonical way.
What is known about the fixpoint algebra of this action? Is it simple? Is it again a UHF algebra?
The answer is more interesting than I thought it would be. First, a clarification: you mean, that $k$ is fixed; the question still makes sense if $k$ is allowed to vary, and the analysis becomes more complicated (but can be done).
Edit: I made some serious (computational) errors in the original version, resulting in qualitative changes.
Just to give an idea of the method, let's do the case $n=2$ (outcome: the fp algebra is never UHF, but it is simple with unique trace). We can rewrite $D \otimes D = \otimes (M_k \otimes M_k)$ (the first $\otimes$ on the right means the infinite tensor product), and the action is just $\otimes \pi$ where $\pi$ is the flip. The flip on $M_k \otimes M_k$ of course is implemented by an order two element of $M_{k^2}$, that is, $\pi = \text{Ad } u$, where $u^2 = I$. It is not only product type, it is xerox (stationary).
Thus the automorphism is of product type, namely $\otimes \text{Ad }\phi$, where $\phi$ is the representation ${\bf Z_2 }\to M_k \otimes M_k$ given by $g \mapsto u$. Thus it comes within the purview of two papers of Handelman and Rossmann ([HR], Illinois J. Math. Volume 29, Issue 1 (1985), 51-95 Actions of compact groups on AF C$^∗$-algebras http://projecteuclid.org/euclid.ijm/1256045841, and the earlier one in the IUMJ).
What is important is the character of $\phi$. This isn't entirely trivial to calculate (as I discovered after the first iteration). The trace of $u$ is $k$. Let $\chi$ denote the character of $\phi$. Then $\chi = {k \choose 2}\chi_r + k \chi_0$ where $\chi_r$, $\chi_0$ are respectively the regular rep char, and the trivial char.
Then from [HR], K$_0$ of both the fixed point and crossed products is given as the direct limit of repeated multiplications by the character $\chi$, $\times\, \chi: R({\bf Z}_2) \to R({\bf Z}_2)$, where the $R(\cdot)$ is the representation ring of the enclosed group, the ring viewed as an ordered ring. Translating to a map $\bf Z^2 \to \bf Z^2$, the matrix is (multn by $\chi$ on $R({\bf Z}_2)$) $$ \begin{pmatrix} {k \choose 2} + k & {k \choose 2} \\ {k \choose 2} & {k \choose 2} + k \\ \end{pmatrix}. $$ [I'm not familiar with Latex, so used ams-tex. The matrix is circulant with top row ${k \choose 2} + k , {k \choose 2}$.]
Its determinant is $k^3\neq 0$, so the rank of K$_0$ is two. Moreover, if $k' = 2k$, then we see that the matrix for $k'$ is the square of that for $k$. (Uniqueness of the trace follows from the matrix being primitive and the system being stationary.)
Now we can formalize this for any $n$. We write the action of $S_n$ as a product type action on $D^{\otimes n} = \otimes (M_k)^{\otimes n}$. Again the action on the little $n$-fold tensor product comes from a representation of $S_n$, call the character of the representation $\chi$. We obtain the ordered K$_0$ groups (actually ordered modules over $R(S_n)$) as the direct limit $\times\, \chi: R(S_n) \to R(S_n)$--this formally inverts $\chi$, so we obtain the ring $R(S_n) [1/\chi]$ (in case $\chi$ is a zero divisor—as occurs if it is the regular representation—this has to be interpreted as first factoring out the annihilator, then inverting).
This will be rank 1 iff $\chi$ is a multiple of the regular representation (which forces $n!$ to divide $k^n$), and otherwise its rank is difficult to determine (at least for me) in general. So we obtain, the fixed point algebra is UHF iff $\chi$ is an integer multiple of the regular representation. Doing the computation even in the case that $n=3$ exhausted my patience, but I'm willing to bet that the thing is never a multiple of the reg rep.
At least, if $n!$ doesn't divide $k^n$, then $\chi$ is not a multiple of the reg rep char, so the fixed point algebra is a simple AF algebra with unique trace, and its K$_0$-group has nonzero infinitesimals (and has rank at most the number of partitions of $n$).
Here is my take on answering this question. It more-or-less provides the same as David's answer, but by somewhat different methods.
1) Let us first address the easier question: As David mentioned in his answer, the fixed point algebra is simple and has a unique trace. A permutation action of this sort is always pointwise strongly outer. In this case, the extension of this dynamical system to the weak closure with respect to the unique trace yields the analogous permutation action on the hyperfinite II${}_{1}$-factor, which is outer. By some well-known results due to Kishimoto, the crossed product is then simple and has a unique trace. Since the fixed point algebra sits inside the crossed product as a corner, we get the same statement for the fixed point algebra by Brown's theorem.
2) Now the harder question: For $n\geq 2$, the fixed point algebra is never UHF. In fact, this holds even if you consider the fixed point algebra of any faithful action of a non-trivial finite group $G$ by tensorial permutations. Here is why:
Let $D$ be a UHF-algebra of infinite type and let $\sigma: G\curvearrowright D^{\otimes n}\cong D$ be a faithful action of a non-trivial finite group via tensorial permutations.
Claim: The $K_0$-group of the fixed point algebra $D^\sigma$ has more than one direct summand.
In fact, we can compute the $K$-theory. As mentioned above, the fixed point algebra is a corner in the simple crossed product, and so it suffices to look at the $K_0$-group of the crossed product. By the main result of [1], the crossed product is $D$-absorbing. In particular, $$ D\rtimes_\sigma G \cong (D\rtimes_\sigma G)\otimes D \cong (D\otimes D)\rtimes_{\sigma\otimes\operatorname{id}_D} G. $$ Now by Proposition 4.5 of [2], the action $\sigma\otimes\operatorname{id}_D: G\curvearrowright D\otimes D$ is homotopic to the trivial action. But this immediately yields a homotopy between $D\rtimes_\sigma G$ and $D\rtimes_{\operatorname{id_D}} G \cong D\otimes C^*(G)$. Since $G$ is a non-trivial finite group, its group algebra $C^*(G)$ is a direct sum of at least 2 matrix algebras (it has finite dimension at least 2 and admits a character). Combining all of this, we get that $D^\sigma$ is $KK$-equivalent to $D\otimes C^*(G)$, which is the direct sum of at least two infinite-dimensional UHF-algebras, and thus its K-theory has at least two direct summands.
3) As David already mentioned, the fixed point algebra should be a simple AF-algebra. Instead of getting into detailed calculations, one could also appeal to classification results in order to see this. Because of the above observations, the fixed point algebra $D^\sigma$ is a separable, nuclear, simple, unital, quasidiagonal and $D$-stable C*-algebra with a unique tracial state. By Theorem 6.1 of [3], it is thus TAF in the sense of Lin. It also satisfies the UCT by the above $KK$-equivalence. By Lin's classification theory of TAF-algebras, the only thing left to show is that some simple AF-algebra has the same ordered $K$-theory as $D\otimes C^*(G)$ - and this should be well-known, if I am not mistaken.
[1] I. Hirshberg, W. Winter: Permutations on strongly self-absorbing C*-algebras, Internat. J. Math., 19(9):1137–1145, 2008. (http://arxiv.org/abs/0708.0213)
[2] I. Hirshberg, N. C. Phillips: Rokhlin dimension: obstructions and permanence properties, http://arxiv.org/abs/1410.6581
[3] H. Matui, Y. Sato: Decomposition rank of UHF-absorbing C*-algebras, Duke Math. J. 163, no. 14 (2014), 2687-2708. (http://arxiv.org/abs/1410.6581)
EDIT: My original argument, as outlined above, has a crucial flaw. Namely, it was recently brought to my attention that two homotopic group actions on the same C*-algebra do not necessarily give rise to two homotopy-equivalent C*-dynamical systems . (The terminology is very confusing here) Apparently, there are examples of homotopic actions of a finite group such that the crossed products are not even $KK$-equivalent.
