action of $O(n,\mathbb{R})$ on ${S}^{n-1}$

Question

Is the action of $O(n,\mathbb{R})$ on ${S}^{n-1}$ transitive? I think this is true as orthogonal matrices are supposed to rotate and keep the length fixed, but how do I prove this?

EDIT: Based on Alexander's comment, if $x = (x_{1}, \ldots, x_{n}) \in S^{n-1}$, then I can construct a matrix $A$ whose first row is the vector $x$ and whose other rows can be obtained by completing it to an orthonormal basis. But does it ensure that columns are also orthogonal?

Answer 1

With the said alexander:Let $x$ be a arbitary point in $S^{n-1}$,we know there exist a orthonormal basis in $\mathbb{R^n},\alpha_1,\alpha_2,...,\alpha_n$ and $\alpha_1=x$,now consider the linear map $T:\mathbb{R^n}\to \mathbb{R^n}$ s,t $e_i\mapsto\alpha_i(e_i$is standard basis which is also orthonormal).

Call the matrix of $T$,$A$.$A$ is orthonormal and$A{(1,0,...,0)}^t=x^t$.here PROPOSITION 10C part(b)

Answer 2

A very closely related recent question, where it is shown that you can rotate any point to the South Pole. As you are not picky about the sign of the determinant let me summarize: Let $A$ and $B$ be two distinct points on the sphere. What happens if you reflect the sphere w.r.t. the (hyper)plane $H$ that has $\vec{AB}$ as its normal and bisects the line segment $AB$?

If $\vec{n}=\vec{AB}$, then the formula for the said reflection is $$ \vec{x}\mapsto \vec{x}-2\frac{\vec{x}\cdot\vec{n}}{\Vert\vec{n}\Vert^2}\,\vec{n}. $$ With $\vec{n}$ known finding the matrix of this reflection is straightforward.