From D.W.'s comment, we see that your problem is equivalent to the same problem with $s=0$ (and with $d$ only increased by 1). The resulting problem is precisely the Linear Code Equivalence Problem over $\mathbb{F}_2$, just phrased slightly differently. (Usually CodeEq is phrased as "you have two $d \times n$ generator matrices of linear codes, is there a permutation of the $n$ columns which makes the two generator matrices have the same row-span?" In your case, the action by $GL(d,2)$ is the action which preserves the row-span, and the fact that you are only interested in a set of vectors means you essentially have an action of $S_n$ on the columns.)
Here's what's known about its complexity:
- This problem is at least as hard as Graph Isomorphism (essentially goes back to Luks '93 "Permutation groups and polynomial-time computation", see Miyazaki, though it is often cited to Petrank-Roth '97)
- As with GI, it is in $\mathsf{NP} \cap \mathsf{coAM} \cap \mathsf{SZK}$, by essentially the same proof as for GI.
- It's no harder than Tensor Isomorphism [GQ19]
- The best known worst-case upper bound is $2^{O(n)}$ (due to Babai, cf. [BCGQ11])
- There is an algorithm by Sendrier 2000 that may work well in practice, depending on what kinds of codes you are looking at.