I've noticed that the group $\mathbb{Z_n}$ under multiplication has a non-trivial idempotent element particularly when $n=2(2k+1), k \in \mathbb{N}$ --- in other words, the even numbers are twice an odd number.
Claim: For $\mathbb{Z}_n$ and $n=2(2k+1), k \in \mathbb{N}$, $2k+1$ is idempotent
Proof
Let $n=2(2k+1)$ for some $k \in \mathbb{N}$ We want to show that $(2k+1)^2 = 2k+1$ in $\mathbb{Z}_n$.
Since $n = 0 $ in $\mathbb{Z}_n$ we have $2(2k+1) = 0 \iff 4k+2 = 0 \iff 4k = -2$
So $4k = -2$.
Then, $$(2k+1)^2=4k^2+4k+1=k(4k)+4k+1 =k(-2)+4k+1=-2k+4k+1=2k+1$$
Thus, we have shown that $2k+1$ is idempotent in $\mathbb{Z}_n$ under multiplication $\hspace{1cm} \square$
