Given $\{x_i\}_{1 \leq i \leq m}$ and $\{y_i\}_{1 \leq i \leq m}$ such that $x_i \cdot x_j$ = $y_i \cdot y_j, \forall 1 \leq i,j \leq m$, what can I conclude about the two sets of vectors?
Clearly they need not be identical as $y_i = Ax_i$ for any orthogonal matrix $A$ is a valid solution. Intuitively it seems that all valid solutions may be of this form, but I am not certain that is the case, and would ideally like an algebraic proof if it is the case.
As a follow on question, I am also curious if the answer would change if instead we said $x_i \cdot x_j$ = $y_i \cdot y_j, \forall 1 \leq i \neq j \leq m$, so that we no longer had (explicitly anyway) that $\lVert x_i \rVert$ = $\lVert y_i \rVert$.
Any help would be much appreciated, thanks.
The following holds under the assumption of the problem: $x_i\cdot x_j=y_i\cdot y_j$ for all $1\leq i,j\leq m$.
Lemma. If $c_1x_1+\dots+c_mx_m=0$, then also $c_1y_1+\dots+c_my_m=0$.
Proof. $$ \begin{split} \|c_1y_1+\dots+c_my_m\|^2 &= \sum_{1\leq i,j\leq m} c_ic_j y_i\cdot y_j \\ &= \sum_{1\leq i,j\leq m} c_ic_j x_i\cdot x_j = \|c_1x_1+\dots+c_mx_m\|^2 = 0 . \end{split} $$ □
Now I define a map $A:\mathrm{span}\{x_1,\dots,x_m\}\to\mathrm{span}\{y_1,\dots,y_m\}$.
For $x=a_1x_1+\dots+a_mx_m\in\mathrm{span}\{x_1,\dots,x_m\}$, define $$ Ax = a_1y_1+\dots+a_my_m. $$ The map is well defined because if $$ x=a_1x_1+\dots+a_mx_m=b_1x_1+\dots+b_mx_m, $$ then letting $c_i=a_i-b_i$ we have $c_1x_1+\dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+\dots+c_my_m$, which is equivalent to $$ a_1y_1+\dots+a_my_m = b_1y_1+\dots+b_my_m. $$
The map is clearly linear. Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+\dots+a_mx_m$, then $$ \begin{split} \|Ax\| &= \|a_1y_1+\dots+a_my_m\|^2 = \sum_{1\leq i,j\leq m} a_ia_j y_i\cdot y_j \\ &= \sum_{1\leq i,j\leq m} a_ia_j x_i\cdot x_j = \|a_1x_1+\dots+a_mx_m\|^2 = \|x\|^2 . \end{split} $$
This means that $A$ is orthogonal on its domain of definition, which is $\mathrm{span}\{x_1,\dots,x_m\}$, and can therefore be completed to an orthogonal linear transformation $A:\mathbb R^n\to\mathbb R^n$.
Notice that $A$ is uniquely determined only on $\mathrm{span}\{x_1,\dots,x_m\}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.