Getting all permutations of vertices of a regular $n$-simplex through rotations (plus a reflection). Suppose that, starting from a regular $n$-simplex with vertices labeled with $(v_1, v_2, \ldots v_n)$ I want to obtain an arbitrary permutation of the vertices $(w_1, w_2, \ldots w_n)$ out of the $n!$ possible, is it enough to:

*

*Rotate the $n$-simplex around its centroid till $w_i$ is in the position occupied by $v_i$, $i = 1, \ldots ,n$ in the original positioning of the $n$-simplex;

*If step 1. is not enough, exchange $v_1$ and $v_2$ in the original positioning and then repeat step 1. till $w_1$ replaces $v_2$, $w_2$ replaces $v_1$, and $w_i$ is in the position occupied by $v_i$, $i = 3, \ldots ,n$ in the original orientation of the $n$-simplex;

Regarding the rotation about the centroid, for 2D it is clear, for 3D I mean any sequence of single rotations of any angle around whatever axis passing by the centroid. For 4D or more, something similar... What I mean is I can move the solid provided that its integrity is preserved and the centroid stay still. After comments and answers, a more formal way to say this is apply a rigid motion fixing the centroid, or apply an orientation preserving isometry of the ambient space fixing the centroid.
Is this feasible?
 A: If by

Rotate the $n$-simplex around its centroid

You mean "apply an orientation preserving isometry fixing the centroid", then the answer is yes. One way to see this is that $S_n$ is isomorphic to the Coxeter group of type $A_{n-1}$. In particular it is generated by the reflections $s_1,\dots,s_{n-1}$ which algebraically correspond to the transpositions $s_i=(i\;i+1)$ for $1\le i\le n-1$, and geometrically correspond to the reflections of the $n$-simplex in the hyperplane containing the points $v_1,\dots,v_{i-1},v_{i+2}\dots v_n$ and the midpoint of the edge joining $v_i$ and $v_{i+1}$.
Now any permutation of the vertices can be achieved by an element of $S_n$, and eny element of $S_n$ can be written as a product $t_1\cdots t_k$ for $t_i\in\{s_1,\dots,s_{n-1}\}$. If $k$ is even this is a product of an even number of orientation reversing isometries (reflections), which is an orientation preserving isometry (case 1 in the question). If $k$ is odd, $t_1$ is a reflections, and $t_2\cdots t_k$ is an orientation preserving isometry (case 2 in the question).
Actually case 2 specifically askes that $t_1=s_1=(1\;2)$, to achieve this when $t_1\neq s_1$, pre-compose $t_1\cdots t_k$ by $s_1s_1$, the identity.
Note that in up to three dimensions, all orientation preserving isometries fixing a point are rotations about an axis (or the identity). In higher dimensions, there can be other orientation preserving isometries, so if specifically only rotations are allowed in higher dimensions, then the question is not possible.
A: The most straightforward thing to do is a sequence of up to $\binom n2$ reflections.
Assuming a regular $n$-simplex, we can draw a hyperplane through $n-2$ of the points, passing through the midpoint of the edge between the remaining $2$ points, and perpendicular to that edge. A reflection through that hyperplane swaps those $2$ points while keeping the rest fixed.
Since we can generate any permutation by a sequence of swaps, we can permute the vertices however we like by a sequence of reflections.
Indirectly, this method also confirms that a plan like the one you describe also works. The composition of two reflections is a rotation (through the $(n-2)$-dimensional "axis" where the two hyperplanes intersect). In this case, all of the hyperplanes above pass through the centroid, so the "axes" will also contain the centroid.
So just take a sequence of reflections that gives you the permutation you want, then group them up into consecutive pairs. This will give you a sequence of rotations, possibly with one reflection left over.
(I should clarify that I mean a simple rotation here. In 4-dimensional space is when this first matters: a "simple rotation" fixes a plane, and "double rotations" are weirder. For higher values of $n$, you get even more complexity, but the composition of two reflections is always a simple rotation.)
