Let $A$ and $B$ be two real $(n+1) \times (n+1)$ matrices such that $AB=BA$. Let $\phi$ be the action
$$\phi\colon \mathbb{R}^2 \times \mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$$
with
$$\phi(s,t,p)=e^{sA+tB} \cdot p.$$
Consider the action restricted to the sphere:
$$\widetilde{\phi}\colon \mathbb{R}^2 \times S^n \to S^n$$
$$\widetilde{\phi}(s,t,p)=\frac{e^{sA+tB} \cdot p}{\|e^{sA+tB} \cdot p\|}.$$
Now fix a point $p \in S^n$. I would like to prove that The orbit $O_p(\widetilde{\phi})$ has dimension 0 or 1 if and only if $A$, $B$ and $I$ (identity matix) are linearly dependent.
I use the fact that $O_p(\widetilde{\phi}) \approx \mathbb{R}^2/G_p$, where $G_p$ is the isotropy group of $p$. Then $\dim O_p(\widetilde{\phi}) = 2 - \dim G_p$. It implies that $\dim O_p(\widetilde{\phi})$ is 0 or 1 if and only if $\dim G_p$ is 1 or 2.
I managed to prove one part: if $A$, $B$, and $I$ are linearly dependent then
$$s_0 A + t_0 B = \lambda I,$$
for some $s_0, t_0, \lambda \in \mathbb{R}$. Note that $(s_0, t_0) \neq (0,0)$.
Then $\{c(s_0, t_0), c \in \mathbb{R}\}$ is a subgroup of $G_p$ and so $\dim G_p \geq 1$.
I had some problems with the other part. This is what I tried to do:
If $\dim G_p$ is 1 or 2, then there exists a $(s_0,t_0) \in G_p$, $(s_0,t_0)\neq (0,0)$, such that $\{c(s_0,t_0), c \in \mathbb{R} \} < G_p$. Then, for all $c \in \mathbb{R}$,
$$\widetilde{\phi}(cs_0,ct_0,p)=p,$$
i.e.,
$$e^{c(s_0A+t_0B)} \cdot p = \|e^{c(s_0A+t_0B)} \cdot p\|p = e^{(\ln \lambda_c) I} \cdot p, $$
where $\lambda_c = \|e^{c(s_0A+t_0B)}p\|$. I thought about using the fact that the matrix exponential is a diffeomorphism in a neighborhood of 0 and then take $c$ sufficiently small so that I could state that
$$c(s_0A+t_0B) = (\ln \lambda_c) I,$$
but I don't think I could do it, since I don't know that $e^{c(s_0A+t_0B)} \cdot p = e^{(\ln \lambda_c) I} \cdot p$ for all points $p \in S^n$. Maybe there is another way to do this.