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)

Here is my take on answering this question. It more-or-less provides the same as David's answer, but by somewhat different methods.

[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.