Suppose we have two self-conjugate sets $E,F$， each consisting of $n$ distinct complex numbers where $E=\{\lambda_1, \dots, \lambda_n\}$ and $F=\{\eta_1, \dots, \eta_n\}$. By self-conjugate I mean $E^{*} = \{\lambda_1^*, \dots, \lambda_n^*\} = \{\lambda_1, \dots, \lambda_n\}=E$. Further assume $E\neq F$, that is, there exists at least one index $j$ such that $\lambda_j \notin F$.
I am wondering whether there exists an ordering on the sets, such that convex combinations of these two ordered sets have distinct components, i.e., if we let $\tilde{E}, \tilde{F}$ be the ordered sets, then $(1-t)E+tF = \{(1-t)\tilde{\lambda_i} + t \tilde{\eta_i}: i=1, \dots, n\}$ has distinct elements for every $t \in [0,1]$. Note here, by convex combination I mean points in the line segment $\tilde{E}$ and $\tilde{F}$ if we identify $\tilde{E}, \tilde{F}$ in $\mathbb C^n$.
I think if we order the set in nonincreasing order by magnitude, $|\tilde{\lambda_1}| \le \dots \le |\tilde{\lambda_n}|$ and if we have equality in the magnitude, we order them by imaginary parts in nonincreasing order, it seems to do the work. I tried several examples and it does seem work. But certainly examples are not exhaustive and I am hoping to have a formal proof.