In a recent preprint arXiv:1811.10503, I proved that if $a_1,\ldots,a_n$ are distinct elements of a torsion-free additive abelian group $G$, then there is a permutation $\pi\in S_n$ such that all those elements $ka_{\pi(k)}\ (k=1,\ldots,n)$ are pairwise distinct. I also showed that for any odd prime $p$ there is no permutation $\pi$ such that all the numbers $k\pi(k)\ (k=1,\ldots,p-1)$ are pairwise incongruent modulo $p$. In view of this, it is natural for me to pose the following new combinatorial problem for finite groups.

**Conjecture**. Let $G$ be a finite multiplicative group with $|G|>1$, and let $p(G)$ be the smallest prime factor of $|G|$. If $A$ is a subset of $G$ with $0<|A|=n<p(G)-1$, then we may write $A=\{a_1,\ldots,a_n\}$ such that all those
$$a_1,\ a_2^2,\ \ldots,\ a_n^n$$
are pairwise distinct.

I have confirmed this for $n\le3$. For $n=4,5,6,7,8,9$, I have verified the conjecture for cyclic groups $G$ with $|G|$ not exceeding $$100,\ 100,\ 70,\ 60,\ 30,\ 30$$ respectively.

I'm confident that my above conjecture should be true (at least for finite abelian groups). The problem looks quite challenging. Any ideas towards its solution?

PS: If we want to extend the conjecture to include infinite groups, then I think it suffices that $G$ has no non-identity element of order not exceeding $n+1$.