Is the sum of discrete sinusoids an eigenfunction of an LTI system?
Given an LTI system with impulse response $h(n)$, a complex sinusoid of the form $e^{j\omega_0n}$ is always an eigenfunction:

$$\mathcal{T}\{ e^{j\omega_0 n} \} = \sum_{k=-\infty}^{\infty} h(k) e^{j \omega_0 (n-k)} = e^{j \omega_0 n} \left( \sum_{k=-\infty}^{\infty} h(k) e^{-j \omega_0 k}\right) = H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n}$$

Note, however, that the sum of eigenfunctions corresponding to different eigenvalues is not an eigenfunction. You can see this question for a demonstration in a generic vector space. In this case it can be easily seen that, because the system is LTI (in particular, linear), the sum of the respective eigenfunctions will appear at the output:

$$\mathcal{T}\{ e^{j\omega_0 n} + e^{j2\omega_0 n} \} = 
H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n} + H\left(e^{j2\omega_0}\right) \cdot e^{j 2\omega_0 n} $$

As you can see, there is no way to express the output as $K\cdot\left(e^{j\omega_0 n} + e^{j2\omega_0 n}\right)$, except in the case where the two eigenvalues (as stated above, $H\left(e^{j\omega_0}\right)$ and $H\left(e^{j2\omega_0}\right)$ in this case) are the same.