Fix some dimension $n \geq 1$ and a number $k \geq 1$ and suppose that $a_1,...,a_k \in \mathbb{R}^n$ are points in $n$-dimensional space such that among all the $nk$ coordinates there is a subset of size at least $nk - \binom{n+1}{2}$ that is algebraically independent (over $\mathbb{Q}$).

Is there an isometry $T$ of $\mathbb{R^n}$ such that the set of the coordinates of $Ta_1,...,Ta_k$ is algebraically independent? In other words, can we "erase" the algebraic dependency using isometries? Note that the group of isometries acting on $\mathbb{R}^n$ has $\binom{n+1}{2}$ degrees of freedom, that is, we can prescribe the position of $\binom{n+1}{2}$ coordinates, so intuitively the conditions are the weakest such that one can still hope that the answer is positive.

This is easy for $n=1$, and in general, it is easy to see how translations affect algebraic dependency. This paper has an explicit construction for $n=2$ (Lemma 3.5), but it seems difficult to generalize this to arbitrary dimension (or maybe it's just that my knowledge of the Euclidean group is lacking). I wonder if there's a high-level approach to this, or indeed, if the interplay between the 'algebraic' and 'geometric' structures on $\mathbb{R}^n$, given by algebraic independence and isometries, respectively, has been studied before.

Fix a dimension $n \geq 1$ and a number $k \geq 1$ and suppose that $a_1,...,a_k \in \mathbb{R}^n$ are points in $n$-dimensional space such that among all the $nk$ coordinates there is a subset of size at least $nk - \binom{n+1}{2}$ that is algebraically independent (over $\mathbb{Q}$).

Is there an isometry $T$ of $\mathbb{R}^n$ such that the set of the coordinates of $Ta_1,...,Ta_k$ is algebraically independent? In other words, can we "erase" the algebraic dependency using isometries? Note that the group of isometries acting on $\mathbb{R}^n$ has $\binom{n+1}{2}$ degrees of freedom, that is, we can prescribe the position of $\binom{n+1}{2}$ coordinates, so intuitively the conditions are the weakest such that one can still hope that the answer is positive.

This is easy for $n=1$, and in general, it is easy to see how translations affect algebraic dependency. This paper has an explicit construction for $n=2$ (Lemma 3.5), but it seems difficult to generalize this to arbitrary dimension (or maybe it's just that my knowledge of the Euclidean group is lacking). I wonder if there's a high-level approach to this, or indeed, if the interplay between the 'algebraic' and 'geometric' structures on $\mathbb{R}^n$, given by algebraic independence and isometries, respectively, has been studied before.