Nontrivial Monodromy of the Universal Stiefel Bundle (and $O(n)$-equivariant vector fields on spheres)

Question

Note: I'm not allowed to embed images into my posts yet, so I've linked my diagrams instead.

Throughout, we will make use of the following result.

Fact. For $H$ a Lie subgroup of $G$, there is a fibration sequence $$G/H \to BH \to BG.$$ The action of $\pi_1(BG) \cong \pi_0(G)$ on $G/H$ is given by left translation. That is, choose an element $g \in G$ to represent the path-component $[g] \in \pi_0(G) \cong \pi_1(BG)$. Then the action is given by $([g], g'H) = gg'H$.

Consider the fibration $$V_{n,2} \to BO(n-2) \to BO(n),$$ where $V_{n,2} = V_2(\mathbb{R}^n)$ is the Stiefel manifold of (orthonormal) $2$-frames in $\mathbb{R}^n$. Assume that $n$ is $>4$ and even.

I desperately need to understand the statement

Proposition. $\pi_1(BO(n))$ acts by negation on the $\mathbb{Z}$-summand of $$\pi_{n-1}(V_{n,2}) \cong \mathbb{Z} \oplus \mathbb{Z}_2.$$ [Stated by Emery Thomas, Atiyah, BJ Pollina, James McClendon, and others]

First, recall that the diagonal action of $O(n)$ on $V_{n,2}$ is given by $h(v_1,v_2) = (h(v_1),h(v_2))$. This gives rise to the identification $$V_{n,2} = O(n)/O(n-2).$$

Second, we know that $\pi_1(BO(n)) \cong \pi_0(BO(n)) = \mathbb{Z}_2$. Let \begin{equation*} g = \begin{pmatrix} -1 & & & \\ & 1 & & \\ & & \ddots & \\ & & &1 \end{pmatrix} \in O(n)\end{equation*} be a nontrivial generator of $\pi_0(BO(n))$.

Then the projection $p \colon V_{n,2} \to S^{n-1}, (v,w) \mapsto w$, and the inclusion of the fiber $i \colon S^{n-2} \to V_{n,2}, v \mapsto (v,(0,\dots, 0,1))$, are equivariant with respect to the action of $g$.

Diagram 1

Additionally, since $n$ is even, the fibration $S^{n-2}\to V_{n,2} \overset{p}\to S^{n-1}$ admits a section $s \colon S^{n-1} \to V_{n,2}$.

Hence, we have the following (not necessarily commutative) diagram of split short exact sequences on the level of $\pi_{n-1}$:

Diagram 2

The diagram commutes only if the section $s \colon S^{n-1} \to V_{n,2}$ is $g$-equivariant. Alternatively, we can choose different sections $s, s' \colon S^{n-1} \to V_{n,2}$ making the square below commute.

Diagram 3

But this will modify the respective splittings $\pi_{n-1}(V_{n,2}) \cong \mathbb{Z} \oplus \mathbb{Z}_2$. Therefore, we require that $s$ is (vertically) homotopic to $s'$.

Question 1. Does there exist a $g$-equivariant section $s \colon S^{n-1} \to V_{n,2}$ for some $g \in O(n)$ with ${\rm{det}}(g) =-1$?

Question 2. Does there exist (vertically) homotopic sections $s,s'$ such that $gs = s'g$? (Note that sections are homotopic iff they are vertically homotopic [Barcus, "Note on Cross Sections Over $CW$-Complexes"]).

Question 3. It is known that every section $s \colon S^{n-1} \to V_{n,2}$ is (vertically) homotopic to a $\mathbb{Z}_2$-equivariant section (where $\mathbb{Z}_2$ acts antipodally) [Zvengrowski, "Skew Linear Vector Fields on Spheres"]. Does there exist a similar result for the $O(n)$ action described above?

If the answer to question 1/2/3 is positive, then the diagram commutes. Finally, we find that $\pi_{n-1}(S^{n-1}) \overset{g}\to \pi_{n-1}(S^{n-1})$ is multiplication by $-1$ using a degree argument:

Diagram 4

Question 4. If the answers to questions 1,2 and 3 and are all negative, is there another way to see the nontrivial monodromy?