Assume all fixed points are on the boundary. For each fixed point you can find an invariant neighborhood. (You can assume that the $Z/nZ$-action fixes some Riemannian metric. Then any $\epsilon$-ball around the fixed point is invariant.) These sets are homeomorphic to a half-ball.
The complement of the union of these half-balls (over all fixed points) is homeomorphic to a closed ball. The action on this closed ball does not have fixed points, contradicting Brouwer's theorem.
EDIT: The argument works because the cyclic group is finite. (The invariant metric is obtained by averaging any metric over the finite group action.) The argument does not work for infinite cyclic groups because there is no invariant metric and indeed in general no invariant neighborhood for a fixed point.