canonical action of symmetric groups on orthogonal groups There is a canonical faithful orthogonal representation of the symmetric group $S_{n+1}$, for $n\geq 1$:
$$
S_{n+1}\to O(n)
$$
given as follows.
(1). I regard $O(n)$ as the isometry group of the unit sphere $S^{n-1}$ in $\mathbb{R}^{n}$.
(2). Let $\Delta^n$ be a regular $n$-simplex embedded in $\mathbb{R}^{n}$ such that
(i). all its edges are of the same length;
(ii). all its $(n+1)$-vertices are in $S^n$;
(iii). its center is the 0.

(3). I observe that any permutation on the $(n+1)$-vertices of $\Delta^n$ can be uniquely extended to an isometry of $S^n$. Hence I get an embedding of $S_{n+1}$ into $O(n)$.
Regarding $O(n)$ as a manifold, we have a canonical action of $S_{n+1}$ on $O(n)$. Hence we have a covering map
$$
O(n)\to O(n)/S_{n+1}.
$$
Let the vector bundle (the action of $S_{n+1}$ on $\mathbb{R}^{n+1}$ is given by permutation of coordinates of $\mathbb{R}^{n+1}$)
$$
\eta: \mathbb{R}^{n+1}\to O(n)\times _{S_{n+1}}\mathbb{R}^{n+1}\to O(n)/S_{n+1}.
$$
Question: I want to know the Stiefel-Whitney class of $w(\eta)$. How to compute it?  
 A: I think the bundle you consider is trivial, because the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ by permutation of coordinates extends to a representation of $O(n)$. I am not completely sure about the first part of the argument, however. 
I think the map $S_{n+1}\to O(n)$ you describe actually comes from the $n$-dimensional standard representation of $S_{n+1}$. Indeed, permutation of coordinates gives a homomorphism $S_{n+1}\to O(n+1)$, but this has actually values in the stabilizer of the vector $v:=(1,\dots,1)$ and that stabilizer is isomorphic to $O(n)$, via the action on $v^\perp$. Now I think that if you orthogonally project the unit vectors to this hyperplane, you get (up to a multiplication by a constant factor depending on $n$) the $n+1$ points in the sphere that you describe. 
If this is correct, then the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ indeed extends to $O(n)$ as the direct sum of the defining representation of $O(n)$ on $v^\perp$ and the trivial representation on the line spanned by $v$. But then it follows from general principles that the bundle $O(n)\times_{S_{n+1}}\mathbb R^{n+1}$ can be trivialized. To do this, consider the map $O(n)\times\mathbb R^{n+1}\to (O(n)/S_{n+1})\times\mathbb R^{n+1}$ defined by $(g,v)\mapsto (gS_{n+1},g\cdot v)$. This evidently factorizes to an isomorphism on the induced bundle.       
