Note by simple geometry, if the $n_i$ is different, then $\alpha$ is strictly inside the unit circle in the complex plane.
But so are all its $\phi(n)$ conjugates in the cyclotomic field, because the conjugates are all also averages of $m$ $n$th roots of unity, not all equal.
So the norm of $\alpha$ in the cyclotomic field has absolute value less than $1$. For $\alpha$ to be an algebraic integer, its norm must be an integer, and hence its norm must be zero. Therefore, $\alpha=0$.
So the only way for $\alpha$ to have non-zero norm is if all the $n_i$ are equal, and then $\alpha=\zeta^{n_i}$.
So the two things you need to know are:
- A specific "strong" convexity condition on the closed unit disk - the average of points on the boundary are strictly inside the unit disk unless all the points are equal.
- The automorphisms of the cyclotomic field send averages of $n$-th roots of unity to averages of $n$-roots of unity.