= Solution
Let $S=\sum_{i=1}^nX_i$. The <likelihood ratio> of rate two to rate one is $2^n e^{-S}$, which decreases strictly with $S$. By the <Neyman-Pearson lemma>, the \b[most powerful test rejects for small values of the sum]:
$$
\boxed{\text{reject if }S<c_\alpha,\qquad \mathbb P_{\lambda=1}(S<c_\alpha)=\alpha.}
$$
Under the null, $S$ has the <gamma distribution> with shape $n$ and rate one. Thus $c_\alpha$ is its $\alpha$-quantile, equivalently half the $\alpha$-quantile of the <chi-squared distribution> with $2n$ degrees of freedom. The <distribution> is continuous, so boundary randomization is unnecessary for $0<\alpha<1$.
For every fixed alternative $\lambda>1$, the <likelihood ratio> against rate one is $\lambda^n e^{-(\lambda-1)S}$, again strictly decreasing. Therefore the same region is most powerful against every such alternative. Moreover, if $G$ has the shape-$n$, rate-one <gamma distribution>, then $S$ at rate $\lambda$ has the same <distribution> as $G/\lambda$. It follows that $\mathbb P_\lambda(S<c_\alpha)=\mathbb P(G<\lambda c_\alpha)$ increases with $\lambda$, so the maximum rejection probability over $\lambda\leq1$ is $\alpha$ at one. Any competing size-$\alpha$ test for the composite null is also of level at most $\alpha$ at rate one, and is bounded in power by the <Neyman-Pearson lemma> for each alternative. \b[The test is uniformly most powerful for the composite hypotheses as well.] This is the <uniformly most powerful test for an exponential rate> based on the sample sum.
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