I want to verify the following diffeomorphism:
Fix $d_{ij}>0$. Let $$N:=\{(x_1,...,x_k)\in \mathbb{R}^{3k}: x_1=0, \|x_i-x_j\|=d_{ij}, 1\leq i<j\leq k \}.$$ If for each $(x_1,...,x_k)\in N$ at least three points $x_i\in \mathbb{R}^{3}$ are not collinear, then $N$ is diffeomorphic to $O(3)$.
My Attempt. First, fix $a=(a_1,...,a_k)\in N$ and choose three vectors $$\hat{a_1}, \hat{a_2}, \hat{a_3}$$ that are not collinear. In this case, all of $\hat{a_i}$'s are not zero (if not, it may be proved the same as the following way). That is, we assume that $\hat{a_i}$'s are linearly independent. In this situation, for each $x=(x_1,...,x_k)\in N$ we may choose three linearly independent vectors $$\hat{x_1},\hat{x_2},\hat{x_3}$$ (that are not collinear) such that the change of basis $$\hat{x_i}=A_x \hat{a_i}\ (i=1,2,3)$$ lies in $SO(3)$ (this may be true but I could not prove it yet). Moreover, then there is some permutation $\sigma_x \in S_k$ such that $$x_{\sigma_x(i)}=A_xa_i\ (i=1,..,k)$$ (this may be true but I could not prove it yet). We define $$\phi : N\rightarrow SO(3)\oplus\{\pm 1\}$$ to be $$\phi (x)=(A_x, \text{sgn}\sigma_x).$$ I have already seen $O(3)=SO(3)\oplus\{\pm 1\}$. I am convinced that the $\phi$ is a diffeomorphism from the view of geometry, but I cannot prove it!
Thank you in advance.