Unique Euclidean isometry between affinely independent points Let $u_0,\dots,u_n$ be vectors in $\mathbb{R}^n$ such that $u_1-u_0,\dots,u_n-u_0$ are linearly independent and similarly let $v_0,\dots,v_n$ be vectors in $\mathbb{R}^n$ such that $v_1-v_0,\dots,v_n-v_0$ are linearly independent. Further suppose that
$$
|u_i-u_j| = |v_i-v_j| \: \text{for all} \: i,j.
$$
I want to show that there is a unique Euclidean isometry $\phi$ which takes $u_1,\dots,u_n$ to $v_1,\dots,v_n$ respectively.
The only idea I've had is to set $\phi(u_i-u_0) = v_i-v_0$ for $i=1,\dots,n$ and then extend linearly to get a linear isomorphism.  But this needn't be distance preserving - if there is to be a unique isometry, I should need to map $u_i-u_0$ to a unique ordering of the $v_i-v_0$ (geometrically, e.g. in $\mathbb{R}^2$, fixing particular vertices $u_0$ and $v_0$ of each triangle, there should be only one way of matching up the pairs of edges coming out from those vertices). But I cannot see any convenient method of finding that ordering, or a method of defining the isometry which is order independent. 
EDIT: The stricken through text is incorrect, for each choice of ordering we get an isometry, it is unique once we fix an ordering and a fixed one is given by the question statement.
Thank you in advance.
 A: First deal with existence:
Let $x_i = u_i -u_0, y_i = v_i-v_0$, and define the linear $Q x_i = y_i$. We have $\|x_i\| = \|Q x_i \|$, and $\|x_i-x_j\| = \|Q (x_i-x_j) \|$ for $i,j =1,...,n$.
Since $\|x_i-x_j\|^2=\|x_i\|^2+\|x_i\|^2-2 \langle x_i, x_j \rangle $, we see that $\langle x_i, x_j \rangle = { 1\over 2} \left( \|x_i\|^2+\|x_i\|^2- \|x_i-x_j\|^2\right)$, from which it follows that $\langle x_i, x_j \rangle = \langle Qx_i, Qx_j \rangle $ for $i,j =1,...,n$, and hence $Q^TQ = I$, so $Q$ is orthogonal.
Now define $\phi(u) = v_0 + Q(u-u_0)$, which is easily seen to be affine and distance preserving.
Now uniqueness:
Suppose $\eta$ is an isometry and $\eta(u_i) = v_i$ for $i,j =0,...,n$. We have $\|\eta(u_i)-\eta(u_j)\|=\|u_i-u_j\|$. Define $A(x)= \eta(x+u_0)-\eta(u_0)$. A little work shows that $A$ preserves the inner product and that $A$ is linear. Consequently, we have $\eta(u) = v_0+A(u-u_0)$, and we see that $\eta$ is affine.
Since $\eta$ is affine and $u_0,...,u_n$ are affinely independent, it follows that $\eta = \phi$, hence it is unique.
