Let $n$ be an odd integer and $a_1, \cdots, a_n$ distinct complex numbers. Suppose that for each $1 \leq i \leq n$, the set $$( \dfrac{a_j}{a_i}: 1 \leq j \leq n, j \neq i )$$ consists of $\dfrac{n-1}{2}$ pair of numbers which are inverse to the other. Can we prove that all of them are equal and whence $n=1$?
I have tried that $n=3$, by the equalities $a^2=bc, b^2=ac$ and $c^2=ab$, we have $a=b=c$. But I do not know that whether this holds in general or not.
Let $F_i=\bigg\{\dfrac{a_j}{a_i}:1\leq j\leq n\bigg\}$. Note that $-1\notin F_i$.
Now, $F_k=\bigg\{\dfrac{a_i}{a_k}z\;\;:\;\;z\in F_i\bigg\}$.
Automatically, we have $\bigg(\dfrac{a_i}{a_k}\bigg)^n=1$. This is because the product of all the elements of $F_k$ should be $1$ and the same product for $F_i$ should also be $1$.
Also, the $a_i$ are distinct, so the set $\{a_i:1\leq i\leq n\}$ is equal to $\{a_1, a_1 w, a_1 w^2,\ldots,a_1 w^{n-1}\}$, where $w=\exp{(2\pi i/n)}$.
So this kind of set is the only kind that could obtain the conditions you want.