Let $S^{n-1} = \{x \in \mathbb{R}^n : x_1^2 + \cdots + x_n^2 = 1\}$. Consider a regular spherical simplex, obtained e.g. by taking a hyperspherical cap, picking $n$ equally-spaced points $P = \{p_1, \dots, p_n\}$ on the boundary (which is isomorphic to $S^{n-2}$), then taking the boundaries of the simplex to be the great (hyper)circles through each $P \setminus \{p_i\}$.
For $n \geq 4$, we know that $\mathbb{R}^{n-1}$ cannot be tiled by congruent regular simplices. Despite this, I see two ways to tile $S^{n-1}$ with congruent regular spherical simplices: (1) the orthants are regular spherical simplices of side length $\pi/2$ (there are $2^n$ of them); (2) inscribing a regular simplex in $S^{n-1}$, then taking the natural partition of the surface tiles $S^{n-1}$ with $n+1$ spherical simplices.
Are there other ways to tile $S^{n-1}$ with congruent regular spherical simplices? Feel free to assume $n \geq 4$.
