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(EDITED) You certainly can't have it for all $\theta$. Consider $n=2$. An eigenfunction $f(\theta_1,\theta_2) = \exp(i k_1 \theta_1 + i k_2 \theta_2)$ for eigenvalue $\lambda = -k_1^2 - k_2^2$ will have $f(0,0) + f(\pi,\pi) = f(0,\pi) + f(\pi,0)$ unless $k_1$ and $k_2$ are both odd (i.e. $\lambda \equiv 2 \mod 4$), in which case $f(0,0) + f(\pi,\pi) = -f(0,\pi) - f(\pi,0)$.

It's certainly not true for all $\theta_1, \ldots, \theta_K$. Thus in the case $d=2$, $f(0,0)+f(\pi,\pi)=f(0,\pi)+f(\pi,0)$.

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