Do Hopf bundles give all relations between these “composition factors”?

Write a fiber bundle $F\to E\to B$ in short as $E=B\ltimes F$ (in analogy with groups).

(This is not necessary, but: given another bundle $X\to B\to Y$, we can write $E=(Y\ltimes X)\ltimes F$, but we may also compose $E\to B$ with $B\to Y$ to get fibrations $E\to Y$, whose fibers $Z$ fit inside fibrations $F\to Z\to X$ hence $Z=X\ltimes F$ and $E=Y\ltimes(X\ltimes F)$ as well. Thus we have a kind of associativity property $(Y\ltimes X)\ltimes F= Y\ltimes(X\ltimes F)$.)

I noticed something about sequences involving $\mathrm{Spin}(7)$.

First of all, unit imaginary octonions are square roots of negative one, so $L:\mathrm{Im}(\mathbb{O})\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$ which sends $u$ to left-multiplication-by-$u$ is "Clifford" and so extends to a representation of the Clifford algebra $\mathrm{Cliff}(\mathrm{Im}(\mathbb{O}))\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$, which restricts to a map $\mathrm{Spin}(7)\to\mathrm{SO}(8)$.

The point-stabilizer of any point in $S^7\subset\mathbb{O}\cong\mathbb{R}^8$ is $G_2=\mathrm{Aut}(\mathbb{O})$, and the point-stabilizer of any point in $S^6\subset\mathrm{Im}(\mathbb{O})$ is $\mathrm{SU}(3)$. Thus, we have some bundles

$$\begin{array}{ccccc} G_2 & \to & \mathrm{Spin}(7) & \to & S^7 \\ \mathrm{SU}(3) & \to & G_2 & \to & S^6 \\ \mathrm{SU}(2) & \to & \mathrm{SU}(3) & \to & S^5 \end{array}$$

which combined with $\mathrm{SU}(2)\simeq S^3$ gives

$$\mathrm{Spin}(7)=S^7\ltimes (S^6\ltimes (S^5\ltimes S^3)),$$

and when combined with the Hopf bundles $S^7=S^4\ltimes S^3$ and $S^3=S^2\ltimes S^1$ becomes

$$\begin{array}{lcl} \mathrm{Spin}(7) & = & (S^4\ltimes (S^2\ltimes S^1))\ltimes (S^6\ltimes(S^5\ltimes S^3)) \\ & \mathrm{or} & (S^4\ltimes S^3)\ltimes(S^6\ltimes(S^5\ltimes(S^2\ltimes S^1))). \end{array}$$

On the other hand, we have $\mathrm{Spin}(n)=S^{n-1}\ltimes\mathrm{Spin}(n-1)$ (where $\mathrm{Spin}(n)\to\mathrm{SO}(n)$ gives an action on $S^{n-1}\subset\mathbb{R}^n$ and we invoke orbit-stabilizer again), which means

$$\mathrm{Spin}(7)=S^6\ltimes(S^5\ltimes(S^4\ltimes (S^3\ltimes (S^2\ltimes S^1)))).$$

So it looks like given any two sets of "composition factors" for $\mathrm{Spin}(7)$, they can made equivalent using the Hopf bundles. Is this a general principle or is there an obvious counterexample? That is, if something can be written as $S^{n_1}\ltimes\cdots\ltimes S^{n_k}$ in two different ways, can the (multisets of) composition factors be equated after using the Hopf bundles to replace $S^7$ with $S^4$ and $S^3$ etc.?

(Hopefully this is not too vague and sloppy so as to be cryptic, and not trivial.)