$\newcommand{\SU}{\mathrm{SU}} \newcommand{\Spin}{\mathrm{Spin}}$There is a fiber sequence $G_2\to \Spin(9) \to \Spin(9)/G_2$, and $G_2$ contains $\SU(3)$ as a subgroup. Is there a space (possibly a topological group) $Y$ which sits in a fiber sequence $\SU(2)\to Y\to \Spin(9)/G_2$, such that there is a canonical map $Y\to \Spin(9)$ which fits into a map of fiber sequences $$\require{AMScd} \begin{CD} \SU(2) @>>> Y @>>> \Spin(9)/G_2\\ @VVV @VVV @| \\ G_2 @>>> \Spin(9) @>>> \Spin(9)/G_2?\\ \end{CD}$$ (I would be happy if such a fiber sequence only existed $2$-locally.) It suffices to show that the classifying map $\Spin(9)/G_2\to BG_2$ lifts through the canonical map $B\SU(2) = \mathbf{H}P^\infty\to BG_2$. The fiber of this map is $G_2/\SU(2)$, but, unfortunately, this isn't a topological group, as far as I can tell. It may be helpful to note that $G_2/\SU(2)$ is the Stiefel manifold $V_2(\mathbf{R}^7)$.
Here is a candidate for $Y$. The space $\Spin(9)/G_2$ is like $\SU(2;\mathbf{O})$, and sits in a fiber sequence $S^7\to \Spin(9)/G_2\to S^{15}$. I believe there is a homotopy pullback square
$$\require{AMScd} \begin{CD} \Spin(9)/G_2 @>>> B\SU(2)\\ @VVV @VVV \\ S^{15} @>>> B\Spin(5), \end{CD}$$
where the map $S^{15}\to B\Spin(5)$ detects the generator(?) of $\pi_{14} \Spin(5) \cong \mathbf{Z}/1680$. Then, the fiber $Y'$ of this map $S^{15}\to B\Spin(5)$ is a candidate for $Y$. I don't know how to construct a map $Y'\to \Spin(9)$.
