1. ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(\mu ,{\sigma }^2)}$, ${\displaystyle Y_1,\cdots ,Y_m\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(\nu ,{\sigma }^2)}$. ${\displaystyle \mu ,\nu ,{\sigma }^2}$ unkown. Consider test ${\displaystyle H_0:\mu =\nu \text{ vs }{H}_1:\mu \ne \nu }$. If we denote ${\displaystyle \theta =\mu -\nu }$, then nuisance parameter here can be ${\displaystyle \lambda =(\mu +\nu ,{\sigma }^2)}$ or ${\displaystyle (\mu ,{\sigma }^2)}$.
2. ${\displaystyle X_1\sim \mathrm{Binomial}(n_1,{\pi }_1),X_2\sim \mathrm{Binomial}(n_2,{\pi }_2)}$. ${\displaystyle H_0:{\pi }_1\le {\pi }_2\text{ vs }{H}_1:{\pi }_1>{\pi }_2}$. Here ${\displaystyle n_1,n_2}$ are known, so they are not nuisance parameters.

Sketch:

1. Any unbiased test has ${\displaystyle \beta ({\theta }_0,\lambda )=\alpha ,\mathrm{\forall }\lambda }$ (by continuity of ${\displaystyle \beta (\theta ,\lambda )}$).
2. If power ${\displaystyle =\alpha }$ on boundary, ${\displaystyle {\mathbb{E}}_{{\theta }_0}[\varphi |U]\stackrel{\mathrm{a}.\mathrm{s}.}{=}\alpha }$. (${\displaystyle U(X)}$ is complete sufficient on boundary submodel).
3. ${\displaystyle {\varphi }^{\ast }}$ optimal among all tests with conditional level ${\displaystyle -\alpha }$ (by reduction to univariate model).

Assume ${\displaystyle \varphi }$ is any unbiased test.
#？ (Need finishing)

${\displaystyle X_i\stackrel{\mathrm{i}.\mathrm{n}.\mathrm{d}}{\sim }\mathrm{Poisson}({\mu }_i),i=1,2}$. ${\displaystyle H_0:{\mu }_1\le {\mu }_2\text{ vs }{H}_1:{\mu }_1>{\mu }_2}$. Then $$p_{\mu }(x)=\prod _{i=1}^2\frac{{\mu }_i^{x_i}{\mathrm{e}}^{-{\mu }_i}}{x_i!}={\mathrm{e}}^{x_1{\eta }_1+x_2{\eta }_2-({\mathrm{e}}^{{\eta }_1}+{\mathrm{e}}^{{\eta }_2})}\frac{1}{x_1!x_2!},$$ where ${\displaystyle {\eta }_i=\log {\mu }_i}$, so test is equivalent to ${\displaystyle H_0:{\eta }_1\le {\eta }_2\text{ vs }{H}_1:{\eta }_1>{\eta }_2}$, and further $$\begin{array}{rl}{p}_{\mu }(x)& ={\mathrm{e}}^{x_1({\eta }_1-{\eta }_2)+(x_1+x_2){\eta }_2-A(\eta )}\frac{1}{x_1!x_2!}\\ & ={\mathrm{e}}^{T(x)\theta +U(x)\lambda -A(\eta )}\frac{1}{x_1!x_2!}\end{array}$$ Denote ${\displaystyle \theta ={\eta }_1-{\eta }_2}$, so test is further equivalent to $$H_0:\theta \le 0\text{ vs }{H}_1:\theta >0.$$ We reject for conditionally large values of ${\displaystyle X_1}$, given ${\displaystyle X_1+X_2=u}$. Since $$\begin{array}{rl}{\mathbb{P}}_{\theta }(X_1=x_1|U=u)& {\propto }_{x_1}{\mathrm{e}}^{x_1\theta }\cdot \frac{u!}{x!(u-x_1)!}\\ & =\mathrm{Binomial}(u,\frac{{\mathrm{e}}^{\theta }}{1+{\mathrm{e}}^{\theta }})\\ & =\mathrm{Binomial}(u,\frac{{\mu }_1}{{\mu }_1+{\mu }_2}),\end{array}$$ because ${\displaystyle {\mathrm{e}}^{\theta }=\frac{{\mu }_1}{{\mu }_2}}$. So in the end we do a binomial test.

${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(\mu ,{\sigma }^2)}$, ${\displaystyle {\sigma }^2>0}$ unknown. ${\displaystyle H_0:\mu =0\text{ vs }{H}_1:\mu \ne 0}$. So $$\begin{array}{rl}{p}_{\mu ,{\sigma }^2}(x)& ={\mathrm{e}}^{\frac{\mu }{{\sigma }^2}\sum _{i=1}^n{X}_i-\frac{1}{2{\sigma }^2}\sum _{i=1}^n{X}_i^2-\frac{n\mu }{2{\sigma }^2}}\cdot {\left(\frac{1}{2\pi {\sigma }^2}\right)}^{\frac{n}{2}}\\ & ={\mathrm{e}}^{\theta T-\lambda U-\frac{n\mu }{2{\sigma }^2}}\cdot {\left(\frac{1}{2\pi {\sigma }^2}\right)}^{\frac{n}{2}},\end{array}$$ here ${\displaystyle T=\bar{X},U=||X|{|}^2}$. Optimal tests rejects when ${\displaystyle \bar{X}}$ is extreme given ${\displaystyle ||X|{|}^2}$.
If ${\displaystyle \mu =0}$, ${\displaystyle p}$ is rotationally symmetric, so $$X\mid ||X|{|}^2=\mu \stackrel{H_0}{\sim }\mathrm{Uniform}(\sqrt{n}\cdot s^{n-1}),$$ i.e. $$\frac{X}{||X||}\stackrel{H_0}{\sim }\mathrm{Uniform}(s^{n-1})\perp \perp ||X||.$$ So optimal test rejects when ${\displaystyle \frac{\bar{X}}{||X||}}$ is extreme.

Geometric picture for ${\displaystyle n=2}$:

${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }P,Y_1,\cdots ,Y_m\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }Q}$. ${\displaystyle H_0:P=Q\text{ vs }{H}_1:P\ne Q}$. Under ${\displaystyle H_0}$, ${\displaystyle P=Q}$, ${\displaystyle X_1,\cdots ,X_n,Y_1,\cdots ,Y_m\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }P}$. Let ${\displaystyle (Z_1,\cdots ,Z_{n+m})=(X_1,\cdots ,X_n,Y_1,\cdots ,Y_m)}$. Under ${\displaystyle H_0}$, ${\displaystyle U(Z)=(Z_{(1)},\cdots ,Z_{(n+m)})}$ is complete sufficient.
Let ${\displaystyle S_{n+m}=\{\text{Permutations on }n+m\text{ elements}\}}$. Then $$(X,Y)|U\stackrel{H_0}{\sim }\mathrm{Uniform}(\{\pi U:\pi \in S_{n+m}\}).$$
Thus for any test statistic ${\displaystyle T}$, if ${\displaystyle P=Q}$, $${\mathbb{P}}_{P,Q}(T(z)\ge t|u)=\frac{1}{(n+m)!}\sum _{\pi \in S_{n+m}}\mathbf{1}\{T(\pi Z)\ge t\}.$$

Monte Carlo test:
In practice, we sample ${\displaystyle {\pi }_1,\cdots ,{\pi }_B\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }{S}_{n+m}}$, e.g. ${\displaystyle B=1000}$. Then ${\displaystyle Z,{\pi }_{1}Z,\cdots ,{\pi }_{B}Z\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Uniform}(S_{n+m}(U))}$ under ${\displaystyle H_0}$. The MC p-value is $$\begin{array}{rl}p& =\frac{1}{1+B}(1+\sum _{b=1}^B\mathbf{1}\{T(Z)\le T({\pi }_{b}Z)\})\\ & \stackrel{H_0}{\sim }\mathrm{Uniform}\left(\{\frac{1}{1+B},\cdots ,\frac{B-1}{1+B},1\}\right).\end{array}$$