Let $G$ be a finite non-abelian group of order $n$. Given $g\in G$ we denote its order by $\mathrm{ord}(g)$.

Consider the group algebra $\mathbb{F}[G]$ for some field $\mathbb{F}$.

Given an ordering $g_1,\ldots,g_n$ of all the elements of $G$, consider the product $$ P=(1+g_1+g_1^2+\cdots+g_1^{\mathrm{ord}(g_1)-1})(1+g_2+\cdots+g_2^{\mathrm{ord}(g_2)-1})\cdots (1+g_n+\cdots+g_n^{\mathrm{ord}(g_n)-1}) \in \mathbb{F}[G].$$

Q: Take $\mathbb{F}=\mathbb{C}$. Does there exist an ordering $g_1,\ldots,g_n$ such that $P$ is proportional to $g_1+g_2+\cdots+g_n$?

Here "proportional" means it is equal to $c(g_1+g_2+\cdots+g_n)$ for a scalar $c\in \mathbb{C}$. If the answer is positive then $c$ is determined uniquely , as $\prod_{g\in G}\mathrm{ord}(g)/n$.

Some comments:

1. If $G$ is abelian, $P$ is independent of the ordering and the answer is easily seen to be positive. However, for $G=A_5$, I was able to find (computationally) some orderings for which $P$ is not proportional to $g_1+\ldots+g_{60}$, which is the reason I merely ask for an existence of an ordering.
2. My colleague B. Bedert, who introduced me to group algebras, suggested that the hardest case is when $G$ is a simple non-abelian group.
3. This question is motivated by an old question I asked.
4. See here for a variant of the question which was solved. The solution there suggests that even $n$ might be problematic, so I will also be happy with an answer that restricts to odd $n$.