I can not answer the question but maybe the following perspectives helps clarify the obstruction. let $k \leq n$ and consider
$$
GL(n) \overset{span}{\longrightarrow} Gr(k, n) \overset{\approx}{\longleftrightarrow} O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n)
$$
$GL(n)$ is the group of invertible matrices on $\mathbb{R}^n$, i.e. this represents your iniital bases. $Gr(k, n)$ is the Grassmanian, the differentiable manifold of k-dimensional subspaces of $\mathbb{R}^n$. One way to construct the Grassmanian is to start with the orthogonal group $O(n)$, i.e. the orthogonal matrices on $\mathbb{R}^n$, and take the quotient by the embedding $(O(k) \times O(n - k)) \longrightarrow O(n)$ given by the matrix which applies the $O(k)$ part to the first $k$ dimensions and the $O(n - k)$ part to the remaining $n - k$ dimensions. Basically, an element of $O(n) / (O(k) \times O(n - k))$ is an equivalence classes of orthogonal matrices on $\mathbb{R}^n$ which leave invariant some $k$-dimensional subspace. What would be natural to do is to invert the projection $\pi_{n,k}$, so that $\pi_{n,k}^{-1} \circ \approx \circ span$ gives you your desired map. Of course $\pi_{n,k}$ is not injective, so you can not invert it. But maybe one can study this map and find an atlas of natural subsets and a natural choice of $\pi_{n,k}^{-1}$ somehow.