This is problem 1.9 of chapter 2 from Lehmann and Casella:
Let $X_1, \dots, X_n$ be a sample (assuming i.i.d.) from a discrete distribution which assigns probability $1/3$ to each of the points $\theta-1, \theta, \theta+1$. $\theta$ ranges over all integers. Prove that no non-constant function of $\theta$ has a UMVUE.
My approach: Assume the simplest case $n=1$. We can use the relationship between UMVUE and unbiased estimator of zero to tackle this issue.
Suppose $T(x)$ is a UMVUE for $g(\theta)$ for some function $g$. Then for an unbiased estimator of zero $U(x)$, we have $$ E_\theta U = 1/3(U(\theta-1) + U(\theta) + U(\theta+1)) = 0, \ \forall \theta \in \mathbb{Z}, $$ i.e., $$ U(\theta-1) + U(\theta) + U(\theta+1) = 0, \ \forall \theta \in \mathbb{Z}. $$ Then we have $$ U(\theta-1) = U(\theta + 2), \ \forall \theta \in \mathbb{Z}. $$ Now we use the fact that if $T$ is UMVUE, then for any $U$ that is an unbiased estimator of zero, we have $$ E_\theta[TU] = 0. $$ This implies that $TU$ is also an unbiased estimator of zero. Hence, for any unbiased estimator of zero $U$: $$ T(\theta-1)U(\theta-1) = T(\theta + 2)U(\theta+2), \ \forall \theta \in \mathbb{Z}. $$ This means that $$ T(\theta-1) = T(\theta+2), \ \forall \theta \in \mathbb{Z}. $$ This gives us $$ 1/3(T(\theta-1) + T(\theta) + T(\theta+1)) = 1/3(T(\theta) + T(\theta+1) + T(\theta+2)), \ \forall \theta \in \mathbb{Z}. $$ Now the RHS is $g(\theta)$ while the LHS is $g(\theta+1)$. Therefore, $$ g(\theta) = g(\theta + 1), \ \forall \theta \in \mathbb{Z}. $$ This shows that $g$ is a constant function of $\theta$. That is, UMVUE does not exist for non-constant function of $\theta$.
Now the part I am stuck at is how do I prove this for $n > 1$? I don't see an immediate induction-type argument or something else and carrying out direct derivation through brute force seems extremely complicated.
Any hint would be helpful. Thank you very much!