When can we continuously extend a linearly independent set to a basis? As we all know, any linearly independent subset of $\Bbb{R}^n$ (or vector spaces in general) can be extended to a (Hamel) basis. I'm wondering when this can be done continuously. To put it precisely:

For which $m, n \in \Bbb{N}$, where $m < n$, does there exist a continuous map $\phi$ from the set of real $m \times n$ rectangular matrices of full row rank, to $\operatorname{GL}_n(\Bbb{R})$, such that the submatrix made from the first $m$ rows of $\phi(A)$ is always $A$?

My work so far:
This is a generalisation of an older question that someone else asked, which I answered (twice). One of my answers shows that this cannot be done when $n$ is odd and $m = 1$, using the hairy ball theorem. If such a map existed, then restriction to any row but the first would produce a continuous function mapping a non-zero vector to a linearly independent vector, which contradicts the hairy ball theorem.
Of course, this doesn't work so well when $n$ is even. If $n$ is even, we can simply map
$$(v_1, v_2, \ldots, v_{n-1}, v_n) \mapsto (v_2, -v_1, \ldots, v_n, -v_{n-1}),$$
i.e. swapping pairs of coordinates and negating one. As such, we always get a vector orthogonal to the original vector, and the map is clearly continuous. And, of course, one could simply pick other pairings as well, to get other vectors orthogonal to the original vector. However, there's no guarantee that we could build a whole basis from these maps!
If $m = n - 1$, then such a map exists. Indeed, all we need is a continuous map that takes $A$ and produces a vector linearly independent of all the rows. We can do that by generalising the cross product. We define:
$$\psi(A) = \det\begin{pmatrix} \matrix{e_1 & e_2 & \ldots & e_n} \\ A \end{pmatrix},$$
where $e_1, \ldots, e_n$ are the standard basis vectors for $\Bbb{R}^n$. We compute the "determinant" in much the same way as the usual mnemonic for the cross product:
$$u \times v = \det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix} = \det\pmatrix{u_2 & u_3 \\ v_2 & v_3}\hat{i} - \det\pmatrix{u_1 & u_3 \\ v_1 & v_3}\hat{j} + \det\pmatrix{u_1 & u_2 \\ v_1 & v_2}\hat{k}.$$
This $\psi$ works in much the same way: taking $\psi(A) \cdot v$ for some $v \in \Bbb{R}^n$, the result becomes $\det \pmatrix{v \\ A}$, so if $v$ is already a row of $A$, the result is $0$. Thus, $\psi(A)$ is orthogonal to the rowspace of $A$, and hence adding it as a row to $A$ will produce an invertible square matrix. Given the determinant is continuous, this is also a continuous choice.
Searching yielded some other related posts, but both start with a one parameter family of vectors/rectnagular matrices, rather than considering a function on the full set of matrices. As such, they are asking something weaker.
So, is there some characterisation of when such a map exists?
 A: This is not an answer, rather it is a set of comments too big for the comment field.

The first comment is that the problem can be reformulated in terms of orthonormal sets.  That is, the problem is
equivalent to the following two alternative statements.  As above, we fix integers $m$ and $n$, with $m<n$.

*

*For every ordered $m$-tuple $(e_1, \ldots , e_m)$ of orthonormal  vectors in ${\mathbb R}^n$, there is a Hamel basis
$$
  (e_1, \ldots , e_m, v_{m+1}, \ldots , v_n)
  $$
in which the last $n-m$ vectors depend continuously on the first $m$.


*For every ordered $m$-tuple $(e_1, \ldots , e_m)$ of orthonormal  vectors in ${\mathbb R}^n$, there is an orthonormal basis
$$
  (e_1, \ldots , e_m, e_{m+1}, \ldots , e_n)
  $$
in which the last $n-m$ vectors depend continuously on the first $m$.
Should anyone be curious,  I'd be glad to provide a proof of the above equivalences.

The second comment is that the problem can be reformulated in terms of the triviality of vector bundles.
To be precise,
consider the Stiefel manifold $V_k({\mathbb R}^n)$, namely the set of all orthonormal $k$-frames in ${\mathbb R}^n$ (here I am switching
the notation from $m$ to $k$, as the latter is more standard).  For every $e=(e_1, \ldots , e_k)\in V_k({\mathbb R}^n)$, consider the
subspace $F_e\subseteq {\mathbb R}^n$ given by
$$
  F_e=\{e_1, \ldots , e_k\}^\perp.
  $$
We may then consider the subset of $V_k({\mathbb R}^n)\times {\mathbb R}^n$ given by
$$
  {\scr{F}}=\{(e, v): e\in V_k({\mathbb R}^n),\ v\in F_e\}.
  $$
With the induced topology, $\scr F$ becomes a vector bundle over $V_k({\mathbb R}^n)$, and the question posed by the OP turns out
to be equivalent to whether or not $\scr F$  is a trivial vector bundle. Again  I'd be glad to provide a proof of this
upon request.
