$\newcommand{\S}{\mathbb{S}^1}$ $\newcommand{\la}{\lambda}$
While solving an optimization problem, I reached the following question:
Let $x_1,x_2,x_3,x_4 \in \S$ be four distinct points on the unit circle.
Suppose that there exist strictly positive real numbers $\la_{ij}=\la_{ji}, 1\le i\le j\le 4$ such that $$ \sum_{j \neq i} \la_{ij}x_j \in \text{span}\{x_i\} $$ for every $i \in \{1,2,3,4\}$.
Question: Are the $x_i$ the vertices of a rectangle?
It suffices to prove that $\sum_i x_i=0$.
A rectangle does satisfy the requirement, with $\la_{ij}=1$; Note that $x_2=-x_4, x_1=-x_3$ are antipodal.
Edit:
If we omit the symmetric condition $\la_{ij}=\la_{ji}$ and the positivity condition $\lambda_{ij}>0$, then any "non-degenerate" configuration satisfies this; if any three of the vertices are linearly independent, then we can choose the (not necessarily symmetric coefficients) that satisfy the requirement. But with the symmetric condition this is not so clear.
The answer is no. One counterexample is $$ x_1=(-1,0),\; x_2=\bigl(\tfrac12,\tfrac{\sqrt3}2\bigr),\; x_3=(1,0),\; x_4=\bigl(\tfrac12,-\tfrac{\sqrt3}2\bigr), $$ for which $$ \lambda_{12}=\lambda_{14}=2,\; \lambda_{23}=\lambda_{24}=\lambda_{34}=1 $$ works (for any positive $\lambda_{13}$—this coefficient never matters if the respective points are antipodal).