= Solution
Since $\mathbb P(X_1=0)=e^{-\lambda}$, the indicator $Y$ is an <unbiased estimator> of $\theta$. The joint <probability mass function> is
$$
p_\lambda(x_1,\ldots,x_n)=e^{-n\lambda}\frac{\lambda^{\sum_i x_i}}{\prod_i x_i!}.
$$
The <Fisher-Neyman factorization theorem> therefore makes $T=\sum_iX_i$ a <sufficient statistic>. More explicitly, conditioning on $T=t$ gives a <multinomial distribution> with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the <binomial distribution> $\operatorname{Bin}(t,1/n)$. Consequently conditioning by the <Rao-Blackwell theorem> gives
$$
\boxed{\mathbb E[Y\mid T]=\left(1-\frac1n\right)^T}.
$$
For $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda=0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda>0$, the reduction in <variance> is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the <probability generating function> of $T\sim\operatorname{Poisson}(n\lambda)$, the new <variance> is $e^{-2\lambda}(e^{\lambda/n}-1)$.
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