$\mathcal{O}=\{\{v_1,...,v_n\}\subset \mathbb{R}^n \mid\{v_1,...,v_n\} \text{ is an orthonormal Basis of } \mathbb{R}^n\}$
$\mathcal{S}^{n-1}=\{x\in \mathbb{R}^n \mid \| x\|=1\}$
I have to prove that $\operatorname{O}(n)\times \mathcal{O}\to \mathcal{O}, (A,\{v_1,...,v_n\})\mapsto A\boldsymbol{\cdot} \{v_1,...,v_n\}=\{Av_1,...,Av_n\}$ is a transitive group operation, hence for two elements $\{a_1,...,a_n\},\{b_1,...,b_n\}$ there has to be a $A\in \operatorname{O}(n)$ such that $A\{a_1,...,a_n\}=\{Aa_1,...,Aa_n\}=\{b_1,...,b_n\}$
I don't really know what to prove here. Since $A$ is orthogonal matrix, we know that $|\det(A)|=1\neq 0$ and thus there will be an inverse $A^{-1}$. Isn't this basically a system of equations $Aa_1=b_1,...,Aa_n=b_n$?
Hint:
You want to show for any two orthonormal bases $\{ a_1, \ldots, a_n \}$ and $\{ b_1, \ldots, b_n \}$ there is an orthogonal matrix going from one basis to the other. That is, $A a_k = b_k$ for each $k$.
Of course, since the $\{ a_k \}$ form a basis, you've now defined a map on a basis. So this extends to a linear map in a unique way. Do you know how to find the matrix associated to a linear map? Once you have this matrix in hand, can you show it's orgthogonal? This will amount to checking that each $b_i \cdot b_j = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}$ (do you see why?), which will be true because of the orthonormal-ness of the $\{ b_k \}$ basis.
I hope this helps ^_^