What is the UMVUE of $\exp(-2\lambda)$ for X is a random variable with poisson distribution~poisson $(\lambda)$ If $X_1,X_2,…,X_n\sim Pois(λ)$, find the UMVUE of $\exp(−2λ)$.
I know based on lehmann scheffe theorem, the step is
(1)find $q(x)$ an unbiased estimater of $\exp(−2λ)$
(2) $T(X)$ is sufficient and complete
(3) $s(*)=\mathbb E(q(x)|t(x))$ is a UMVUE
(4) var$(s(*))$ <$\infty$, then it is the UMVUE
The answer IS: y=(-1)^x, when x is even y=1,when x is odd, y=-1.
But I still do not know step by step, how to make it work?
 A: An easier approach is to take the approach in the two linked questions, using an initial unbiased estimator of $\exp(−2\lambda)$ to be $1$ if $X_1+X_2=0$ and to be $0$ if $X_1+X_2 \gt 0$.
Here is a sketch of an approach using an initial unbiased estimator of $\exp(−2\lambda)$ to be $+1$ if $X_1$ is even and to be $-1$ if $X_1$ is odd, which following your notation we might call $q(X_1)$:
1) $q(X_1)$ is an unbiased estimator as its conditional expectation, given $\exp(−2\lambda)$ and so $\lambda$, is $+1\left(e^{-\lambda}+e^{-\lambda}\frac{\lambda^2}{2!}+\cdots\right) -1\left(e^{-\lambda}\frac{\lambda^1}{1!}+e^{-\lambda}\frac{\lambda^3}{3!}+\cdots\right) = e^{-\lambda}\left(1-\frac{\lambda^1}{1!}+\frac{\lambda^2}{2!}-\frac{\lambda^3}{3!}+\cdots\right)=e^{-2\lambda} $
2) A complete sufficient statistic for a Poisson distribution is $T(\mathbf X)=\sum\limits_1^n X_i$
3) The conditional distribution of $X_1$, given $\sum\limits_1^n X_i=t$, is binomial with parameters $t$ and $\frac{1}{n}$, so you can show the conditional probability that $X_1$ is even is $\frac{1+\left(1-\frac2n\right)^t}{2}$ and that $X_1$ is odd is $\frac{1-\left(1-\frac2n\right)^t}{2}$, and thus $\mathbb E[q(X_1) \mid T(\mathbf X)=t] = \left(1-\frac2n\right)^t$ and this is then the uniformly minimum-variance unbiased estimator for $\exp(−2\lambda)$ using the Lehmann–Scheffé theorem
4) This UMVUE estimator is bounded: when $n=1$ it is  $-1$ or $+1$, while for $n \ge 2$ it is bounded by $0$ and $1$. So it has a finite variance. 
