The answer is no.
Take anyA rational edge length is an algebraic dependence between the coordinates that you cannot get rid of by an isometry. One can find $n$$k$ points in $a_1, \ldots, a_n \in \mathbb{R}^n$$\mathbb{R}^n$ with all but one coordinate algebraically independent coordinatesand $\sum_{i=1}^n (a_1^i - a_2^i)^2 = 1$. Choose any point (For this, take $a_{n+1} \in \mathbb{R}^n$$k$ generic points with $a_1$ and $a_2$ at distance less than $1$ from $a_1$. Then there is an algebraic dependence over $\mathbb{Q}$ between, and then change one of the coordinates of $T(a_1)$ and$a_2$ so that the distance from $T(a_{n+1})$ for any isometry$a_1$ becomes $T$$1$.)