Let $a_1, a_2, a_3,\ldots,a_n \in\mathbb R^n$, $a_i = (a_{i1},a_{i2},\ldots,a_{in})$ , $\langle a_i,a_i\rangle = \sum_j a_{ij}a_{ij} = 1$ and if $i \neq k$ then $\langle a_i,a_k\rangle = \sum_j a_{ij}a_{kj} = 0$. If $T$ is a function from $\mathbb R^n$ to $\mathbb R^n$ defined by $T(x_1,x_2,\ldots,x_n) = (y_1, y_2,\ldots,y_n)$ where $y_i = \sum_j a_{ji}x_j$. Then $T$ is an isometry.
\begin{align} d(T(x),T(y)) & = \sqrt{\sum_i \left(\sum_j a_{ji}x_j-a_{ji}y_j \right)^2} \\[10pt] = & \sqrt{\sum_i \left( \sum_j \sum_k a_{ji}a_{ki}(x_j-y_j)(x_k-y_k)\right)} = \sqrt{\sum_i^{} (x_i-y_i)^2} = d(x,y). \end{align}
I did prove that $d(x,y) = d(T(x),(T(y))$ for any $x,y \in\mathbb R^n$. How can i prove that $T$ is surjective?