Let $M$ be a (closed, smooth) manifold of dimension $d$. For $n$ a positive integer, fix $n$ points $x\_1, \dots, x\_n \in M$. The group of diffeomorphisms of $M$ that permutes the points $x\_i$ surjects to the symmetric group $\Sigma\_n$, at least for $d \geq 2$:


$$\Phi\colon \text{Diff}(M,\{x\_1, \dots, x\_n\}) \to \Sigma\_n$$


Is it always true that $\Phi$ does not have a section for $n$ big enough? 


How big do we have to choose $n$ for $M = S^d$? (The naive guess would be $d+3$.) 



Realizing symmetric groups by diffeomorphisms
I'm going to reverse the roles of $n$ and $d$ (since otherwise I will screw things up in this answer). If $M^n$ is a connected $n$-dimensional smooth manifold, then there is no section of the map $Diff(M^n,\{x\_1,\ldots,x\_d\}) \rightarrow \Sigma\_d$ if $d \geq n+3$ (nb: you forgot in your question to assume that $M^n$ is connected). Indeed, assume that $\Sigma\_d$ acts by diffeomorphisms on $M^n$ such that the action restricts to the permutation action on $\{x\_1,\ldots,x\_d\}$. The group $\Sigma\_{d-1} \subset \Sigma\_d$ then fixes $x\_d$, so we get a linear action of $\Sigma\_{d-1}$ on the tangent space $T\_{x\_d} M^n = \mathbb{R}^n$. Every element of $\Sigma\_{d-1}$ must act nontrivially on this tangent space; indeed, by averaging we can find a Riemannian metric on $M^n$ that is preserved by the action of $\Sigma\_d$, and an isometry of a connected Riemannian manifold that fixes a point and all tangent vectors at that point must be the identity. But the smallest dimensional vector space on which $S\_{d-1}$ acts faithfully is $(d-2)$-dimensional. We thus deduce that $n \geq d-2$, so $d \leq n+2$, as desired.


