Understanding that a hypothesis test does not depend on a certain parameter I have a very big doubt regarding a test not depending on $\theta_1$. Suppose I have:
$$(X_i)_{i=1}^{n}, X_i \stackrel{iid}{\sim} N(\theta, 1), \hbox{ that is } X_i \sim g(x_i, \theta) = (2\pi)^{-1/2} \exp[\frac{-1}{2}(x_i - \theta)^2]$$
The joint distribution is given by $f(x;\theta) = (2\pi)^{-n/2} \exp[\frac{-1}{2}\sum_{i = 1}^{n}(x_i - \theta)^2] $
Suppose I have the following hypothesis test with $0<\theta_1$, where $\theta_1$ is fixed but arbitrary (remark: always positive)
$$H_0: \theta = 0\quad \hbox{vs}\quad H_1: \theta = \theta_1$$
By Neyman–Pearson theorem, there is some $k>0$ such that
$$\phi(x) = \left\{
  \begin{array}{lr}
    1 & : f(x;0)k  < f(x;\theta_1)  \\
    0 & : f(x;0)k  > f(x;\theta_1)
  \end{array}
\right.$$
is a UMP test in the class of tests with size $\alpha = E_0[\phi(X)]$.  
$$\hbox{I want to understand why this test does not depend on }\theta_1?$$ 
$remark:$ $\theta_1$ is positive and fixed but arbitrary. That is, I want to understand if I want to change the parameter $\theta_1$, I will still get the same test exactly. Notice that:
$$k < \frac{f(x;\theta_1)}{f(x;0)} \Longleftrightarrow \ln(k) < \frac{n}{2}(2 \theta_1 \bar{x} - \theta_1^{2}) $$
In addition, we can see that
\begin{equation} \label{eq1}
\begin{split}
k < \frac{f(x;\theta_1)}{f(x;0)} & \Longleftrightarrow  \frac{\ln(k)}{n} < \frac{1}{2}(2 \theta_1 \bar{x} - \theta_1^{2}) = \theta_1\bar{x} - \frac{\theta_1^{2}}{2} \\
 & \Longleftrightarrow \frac{\ln(k)}{n} + \frac{\theta_1^{2}}{2} < \theta_1\bar{x}\\
& \stackrel{\theta_1 > 0}{\Longleftrightarrow } \frac{\ln(k)}{n\theta_1} + \frac{\theta_1}{2} < \bar{x}\\
&\Longleftrightarrow \sqrt{n}\left[\frac{\ln(k)}{n\theta_1} + \frac{\theta_1}{2}\right] < \sqrt{n}\bar{x}\\
& \Longleftrightarrow \tilde{k}(\theta_1) < \sqrt{n}\bar{x}
\end{split}
\end{equation}
Then  $\phi(x) = \left\{
  \begin{array}{lr}
    1 & : \tilde{k}(\theta_1) < \sqrt{n}\bar{x}  \\
    0 & : \tilde{k}(\theta_1) > \sqrt{n}\bar{x}
  \end{array}
\right.$, with $\alpha = E_0[\phi(X)]$. Notice that $Z = \sqrt{n} \bar{X} \sim N(0,1)$. So, to determine the test well, we have to determine the constant $\tilde{k}(\theta_1)$. We will determine the constant $\tilde{k}(\theta_1)$ forcing the test to be of size $\alpha$. Under the null hypothesis, we have:
\begin{equation} 
\begin{split}
\alpha = E_{0}[\phi(X)] & \Longleftrightarrow  P_{0}[\tilde{k}(\theta_1) < Z] = \alpha \\
 & \Longleftrightarrow P_{0}[Z \leq \tilde{k}(\theta_1) ] = 1 -\alpha \\
& \Longleftrightarrow F_Z (\tilde{k}(\theta_1)| \theta = 0) = 1 -\alpha\\
& \Longleftrightarrow \tilde{k}(\theta_1) = F_{Z}^{-1} ( 1- \alpha| \theta = 0)
\end{split}
\end{equation}
And here is my problem, because the $\tilde{k}(\theta_1)$ depends on the parameter $\theta_1$. For example, suppose $\alpha = 0.01$, we have
$$\tilde{k}(\theta_1) = \sqrt{n}\left[\frac{\ln(k)}{n\theta_1} + \frac{\theta_1}{2}\right] = 2.33$$
In other words, if a take some other $\theta_1^{'}>0$, we have other $\tilde{k}(\theta_1^{'})$. And consequently, I will have another test $\phi^{'}(x) = \left\{
  \begin{array}{lr}
    1 & : \tilde{k}(\theta_1^{'}) < \sqrt{n}\bar{x}  \\
    0 & : \tilde{k}(\theta_1^{'}) > \sqrt{n}\bar{x}
  \end{array}
\right.$
For this purpose, can I adjust the $\tilde{k}(\theta_1)$ by resizing the sample $n$? that is, if I want the $\tilde{k}(\theta_1)$ not to depend on the parameter $\theta_1$, I should just change $n$? but this does not seem to make much sense. Why am I asking this? because in other problems, I really need to vary the parameter $\theta_1 > 0$ and ensure that the test does not depend on $\theta_1$. For example:
$$H_0: \theta \leq 0\quad \hbox{vs}\quad H_1: \theta > 0.$$
 A: Your test a priori defined as
$$ \phi(x) = \left\{
  \begin{array}{lr}
    1 & : f(x;0)k  < f(x;\theta_1)  \\
    0 & : f(x;0)k  > f(x;\theta_1)
  \end{array}
\right.$$ has $k$ depending on $\theta_1$ but its rejection criterion is equivalent to
$$\tilde{k}(\theta_1) := \sqrt{n}\left[\frac{\ln(k)}{n\theta_1} + \frac{\theta_1}{2}\right] >\sqrt{n}\overline{x}$$ 
where $k$ is chosen so that the LHS is $2.33$ for $\alpha=0.01$. The key point to notice is that the for any other $\theta_1>0$, you may choose $k$ so that $\tilde{k}(\theta_1)=2.33$. This is true since the LHS above is monotone/invertible in $k$ for any fixed $\theta_1>0$. Then, the statement you want to make is that "the test with rejection region $\{2.33<\sqrt{n}\overline{x}\}$ is UMP level $\alpha=0.01$ for testing $H_0:\theta\leq 0$ versus $H_1:\theta>0$". This is true precisely because $f(x;\theta_1)/f(x;0)$ is monotone in $\overline{x}$. See the definition of monotone likelihood ratio and the Karlin-Rubin theorem for a general formulation of this.
A: You appear to have stumbled onto a general property of classical hypothesis tests here, which is that the alternative hypothesis only affects the test through the evidentiary ordering it induces, which measures which data outcomes are more conducive to the null hypothesis and which data outcomes are more conducive to the alternative hypothesis.  (This is measured through the test statistic, which in this case is the likelihood-ratio statistic).
In this particular case, any value $\theta_1 > 0$ will induce the same evidentiary ordering in the test.  Specifically, any posited value of this parameter that is greater than zero is going to lead to an evidentiary ordering where higher observed values of the sample mean are more conductive to the alternative hypothesis, and lower observed values of the sample mean are more conducive to the null hypothesis.  Since any such specification leads to the same evidentiary ordering, and since the alternative hypothesis affects a hypothesis test only through its contribution to this ordering, any such specification leads you to the exact same hypothesis test (i.e., the same p-value function).
What you are seeing here is a general property of classical hypothesis testing which I have described in more detail in this related answer.  As you will see in the related answer, two hypothesis tests are equivalent if they have the same null distribution and impose the same evidentiary ordering on the data.  The alternative hypothesis in a classical hypothesis test performs a very limited role --- it contributes only to the evidentiary ordering in the test, and after this it is effectively ignored.
