Existence of an isometry such that $f(A)=A'$, $f(B)=B'$ and $f(C)=C'$ Let $A,B,C,A',B',C'\in \mathbb S^2_r$. If we have that:

*

*$d(A,B) = d(A', B')$

*$d(A,C) = d(A', C')$

*$d(C,B) = d(C', B')$
Does that mean that there exists an isometry $f: \mathbb S^2_r \to \mathbb S^2_r$ such that $f(A)=A'$, $f(B)=B'$ and $f(C)=C'$?
I know that if $A, B, C$ are non-collinear then an isometry with such property is unique in the sense that if $g$ has the same property, then $g=f$ however, in my class, we just talk about the uniqueness of such isometry but we never talked about its existence. I know that it seems very intuitive that such isometry exists, but how can this be proved?
 A: We can do a construction proof from algebraic perspective. Since all isometries of sphere are orthogonal transforms in $\mathbb R^3$, we just need to construct the orthogonal matrix $M$.
Without loss of generality, let's first assume $r=1$, i.e. unit radius sphere.
Then let's think about A,B,C as (column) vectors in 3d space, $v_A,v_B,v_C$. As you defined
$$
d(A,B)=\arccos(v_A^T v_B)\in [0,\pi]
$$
Thus the equidistance conditions translate to
$$
v_A^Tv_B=v^T_{A'}v_{B'}\\
v_A^Tv_C=v^T_{A'}v_{C'}\\
v_C^Tv_B=v^T_{C'}v_{B'}
$$
Let's combine the vectors into a matrix $V=[v_A,v_B,v_C]$, $V'=[v_{A'},v_{B'},v_{C'}]$.
Then in matrix notation, your condition translates to $V^TV=V'^TV'$.
Finally, let's solve the mapping if A,B,C are not colinear, then the $V$ matrix is invertible (full rank).
We can solve the linear mapping $M$, uniquely by
$$
MV=V'\\
M=V'V^{-1}
$$
Easy to show $M$ is an orthogonal matrix,
$$
V'^TV'=V^TM^TMV = V^TV
$$
since $V$ is invertible
$$
V^TM^TMVV^{-1} = M^TM=V^{-T}V^TVV^{-1}=I\\
$$
Thus we uniquely constructed the isometry of the sphere that maps $A,B,C$ to $A'B'C'$.
