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A metrically homogeneous space is a metric space $(X,d)$ such that for all points $p$ and $q$ in $X$, there exists an isometry $f$ such that $f(p) = q$. Does the sphere $S^n$ have this property?

odd dimension for considering as complex. Just take $f(x)=qp^{−1}x$ which would be the generalization of the rotation. If it even?

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Consider the great circle containing both points. Rotate along that circle. You can do this precisely by looking at $S^n$ as a subspace of $\mathbb{R}^{n+1}$ – Brett Frankel Apr 28 '12 at 19:50
up vote 1 down vote accepted

Yes, $S^{n-1}$ is metrically homogeneous: its isometry group $O(n)$ acts transitively. One way to see this is to complete both $\vec p$ and $\vec q$ to orthonormal bases, and use the transition matrix between these bases.

In fact, $S^{n-1}$ is $m$-point homogeneous for every $m\ge 1$, meaning that for any two $m$-tuples $(x_j)$ and $(y_j)$ such that $d(x_i,x_j)=d(y_i,y_j)$ for all $i,j$, there exists an isometry $f$ such that $f(x_j)=y_j$ for all $j$. See Proposition 12 in Metric spaces of curvature $\ge k$ by C. Plaut.

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