Let $x,y\in S^{n-1}$ be two points on the unit sphere in $\mathbb{R}^n$, then an isometry between them should exist which is just rotation of the space, but I'm having issues showing such rotation exists formally.
My idea so far was to first reduce to showing an isometry exist between $e_1$ (being the standard basis vector $(1,0,\dots,0)$) and any other point $y=(y_1,\dots,y_n)$, and to do that I want to find an isometry $A_i$ between $e_1$ and $e_i$ for any $i\in\{2,\dots,n\}$ and then write for $x\in S^{n-1}$
$$f(x) = \sum_{i=1}^{n}y_iA_ix$$
where $A_1=Id$. We would then indeed have that $f(e_i)=y$ but I'm having issues showing this function is an isometry, as given $a,b\in S^{n-1}$,
$$d(f(a),f(b))=\left|\sum_{i=1}^{n}y_iA_ia-\sum_{i=1}^{n}y_iA_ib\right|=\left|\sum_{i=1}^{n}y_iA_i(a-b)\right|$$
where I'm just not sure how to continue..
It was a good idea to reduce to getting an isometry taking the standard basis vector $e_1$ to any specified unit vector $y$. On the other hand, I don't think the rest of your approach will work; your $f$ is unlikely to be an isometry.
I suggest finding (e.g., by Gram-Schmidt algorithm) an orthonormal basis $(b_1,b_2,\dots,b_n)$ whose first vector $b_1$ is the given $y$. (It doesn't matter what the rest of the $b_i$'s are as long as they're orthonormal.) Then the matrix whose columns are the $b_i$'s (in order) defines (with respect to the standard basis of $e_i$'s) an isometry sending $e_1$ to $y$ (and, more generally, sending each $e_i$ to the corresponding $b_i$).