From the equation $ \omega + \frac{1}{\omega} = 1, $ $ \omega^2 + 1 = \omega, $ so \[\omega^2 - \omega + 1 = 0.\]Then $ (\omega + 1)(\omega^2 - \omega + 1) = 0, $ which expands as $ \omega^3 + 1 = 0 $. Hence, $ \omega^3 = -1 $. We divide into cases where $ n $ is of the form $ 3k, $ $ 3k + 1, $ and $ 3k + 2 $. If $ n = 3k, $ then \[\omega^n + \frac{1}{\omega^n} = \omega^{3k} + \frac{1}{\omega^{3k}} = (\omega^3)^k + \frac{1}{(\omega^3)^k} = (-1)^k + \frac{1}{(-1)^k}.\]If $ k $ is even, then this becomes 2, and if $ k $ is odd, then this becomes $ -2 $. If $ n = 3k + 1, $ then \begin{align*} \omega^n + \frac{1}{\omega^n} &= \omega^{3k + 1} + \frac{1}{\omega^{3k + 1}} = (\omega^3)^k \omega + \frac{1}{(\omega^3)^k \omega} \\ &= (-1)^k \omega + \frac{1}{(-1)^k \omega} \\ &= (-1)^k \frac{\omega^2 + 1}{\omega} \\ &= (-1)^k \frac{-\omega}{\omega} \\ &= (-1)^k.\end{align*}This can be $ 1 $ or $ -1 $. And if $ n = 3k + 2, $ then \begin{align*} \omega^n + \frac{1}{\omega^n} &= \omega^{3k + 2} + \frac{1}{\omega^{3k + 2}} = (\omega^3)^k \omega^2 + \frac{1}{(\omega^3)^k \omega^2} \\ &= (-1)^k \omega^2 + \frac{1}{(-1)^k \omega^2} \\ &= (-1)^k \frac{\omega^4 + 1}{\omega^2} \\ &= (-1)^k \frac{-\omega + 1}{\omega^2} \\ &= (-1)^k \frac{-\omega^2}{\omega^2} \\ &= -(-1)^k.\end{align*}This can be $ 1 $ or $ -1 $. Hence, the possible values of $ \omega^n + \frac{1}{\omega^n} $ are $ \boxed{-2,-1,1,2} $.