We all know Gram-Schmidt orthogonalization is highly non-democratic, i.e. performing the Gram-Schmidt procedure on the linearly independent vectors $v_1, v_2, v_3, \cdots , v_n$ will result in a different set of orthonormal vectors then performing it on $v_{i_1}, v_{i_2}, v_{i_3}, \cdots , v_{i_n}$ , where $i_1, i_2, \cdots , i_n$ is a permutation of $1,2 \cdots n$.
I want to know is it possible to distinguish the two orthonormal sets of vectors generated by the Gram-Schmidt procedure using different set of vectors.
I.e., Is it possible to distinguish between the orthonormal vectors generated by
$$v_1, v_2, v_3, \cdots , v_n \quad \text{and} \quad v_{i_1}, v_{i_2}, v_{i_3}, \cdots , v_{i_n}$$
from the orthonormal vectors generated using the set
$$v_1, v_2, \cdots v_{n-k}, w_1, \cdots w_k$$
If you can provide me any method to distinguish between them it would be great. But I am mostly interested in knowing that even theoretically is it possible to do so?
Let start by looking at a linear space of dimension equal to one. I.e. isomorphic to $\mathbb R$ if we're speaking of linear spaces over the field $\mathbb R$.
$\{1\}$ is a linearly independent set of vector(s). As is $\{2\}$.
For both cases, the Gram–Schmidt process leads to $\{1\}$ as the obtained orthonormal basis. And you have no way to know what was the original basis.
To make it short: the Gram–Schmidt process is not one-to-one.