Let $\omega_1,\omega_2,...,\omega_k$ be $n$th roots of unity (not necessarily primitive) and let $q$ be coprime to $n$. If elements of $\omega_j$ have a vanishing sum is it true that their $q$th powers also have a vanishing sum?
I assume not so the next question is "what is the smallest $n$ for which there is a counterexample?" ($q$ and $n$ are of course natural numbers.)
There's no counterexample; the claim is true, as shown below . . .
Fix positive integers $n,q$, with $\gcd(q,n)=1$.
Let $\omega=\exp\bigl(\frac{2i\pi}{n}\bigr)$.
Claim:
If $e_1,...,e_k$ is a sequence of nonnegative integers such that $$\omega^{e_1}+\cdots+\omega^{e_k}=0$$ then the equality $$\omega^{qe_1}+\cdots+\omega^{qe_k}=0$$ also holds.
Proof:
Let $f(x) = x^{e_1}+\cdots+x^{e_k}$, and let $\phi(x)$ be the $n$-th cyclotomic polynomial.
Since $\gcd(q,n)=1$, it follows that $\omega^q$ is a primitive $n$-th root of unity, hence $\phi(\omega^q)=0$. \begin{align*} \text{Then}\;\;&\omega^{e_1}+\cdots+\omega^{e_k}=0\\[4pt] \implies\;&f(\omega)=0\\[4pt] \implies\;&f(x)=\phi(x)g(x),\;\text{for some}\;g\in\mathbb{Q}[x]\\[4pt] \implies\;&f(x^q)=\phi(x^q)g(x^q)\\[4pt] \implies\;&f(\omega^q)=\phi(\omega^q)g(\omega^q)\\[4pt] \implies\;&f(\omega^q)=0\\[4pt] \implies\;&\omega^{qe_1}+\cdots+\omega^{qe_k}=0\\[4pt] \end{align*} as was to be shown.