@fcop's answer talks about the choice of a region such that the power is maximized. In this illustration, I want to point out that even when we have tests that are considered "the most powerful tests" (so, the power isn't an issue here), the rejection region depends on more than just the hypothesis.
I'll use the definition from the Neyman-Pearson Lemma to test $H_0: \theta = \theta_0$ vs. $H_1:\theta=\theta_1$. As pointed out by @fcop, this Lemma holds only in the case of simple hypothesis. However, my purpose is to provide examples to illustrate my point, so my answer is bounded to have a narrower scope.
From the Wikipedia article:
"(...) the likelihood-ratio test which rejects $H_0$ in favour of $H_1$ when
$$\frac {L(x\mid \theta _{0})}{L(x\mid \theta _{1})} \leq \eta$$
where
$$P(\Lambda (X)\leq \eta \mid H_{0})=\alpha$$
is the most powerful test at significance level $\alpha$ for a threshold $\eta$."
And in the proof of this lemma, we can see how the rejection region is defined for this specific case:
"Define the rejection region of the null hypothesis for the NP test as
$R_{NP}=\left\{x:{\frac {L(x\mid \theta _{0})}{L(x\mid \theta _{1})}}\leq \eta \right\}$
where $\eta$ is chosen so that $P(R_{{NP}},\theta _{0})=\alpha \,.$"
Example 1
Now, suppose we have that $X_1,\ldots,X_n \overset{iid}{\sim} \mathcal{N}(\theta,1)$, and we want to test $H_0: \theta = 3$ vs. $H_1:\theta=5$.
So,
$$L(\mathbf{x} \mid \theta) = (2 \pi)^{-n/2} \exp \left\{-\frac{(\sum x_i - \theta)^2}{2} \right\} = (2 \pi)^{-n/2} \exp \left\{-\frac{(\sum x_i - \bar{x})^2 -n(\bar{x}-\theta)^2}{2} \right\}$$
And, calculating $\frac {L(\mathbf{x} \mid \theta _{0})}{L(\mathbf{x}\mid \theta _{1})}$ yields
\begin{align}
\frac {L(\mathbf{x} \mid \theta _{0})}{L(\mathbf{x}\mid \theta _{1})}
= \frac {L(\mathbf{x} \mid 3)}{L(\mathbf{x}\mid 5)}
&= \frac{(2 \pi)^{-n/2} \exp \left\{-\frac{(\sum x_i - \bar{x})^2 -n(\bar{x}-3)^2}{2} \right\}}{(2 \pi)^{-n/2} \exp \left\{-\frac{(\sum x_i - \bar{x})^2 -n(\bar{x}-5)^2}{2} \right\}} \\
&= \exp\left\{- n(\bar{x}-3)^2 +n(\bar{x}-5)^2 \right\} \\
&= \exp\left\{ -n(4 \bar{x}-16)\right\}
\end{align}
Therefore, the rejection region is defined in terms of
\begin{align}
\frac {L(\mathbf{x} \mid \theta _{0})}{L(\mathbf{x}\mid \theta _{1})} = \exp\left\{ -n(4 \bar{x}-16)\right\} \leq \eta
\end{align}
Simplifying it, we have that
\begin{align}
\exp \left\{ -n(4 \bar{x}-16)\right\} \leq \eta &\iff -n(4 \bar{x}-16) \leq \log(\eta) \\
&\iff \bar{x} \geq \underbrace{4 - \frac{\log(\eta)}{4 n}}_{\mbox{call this }\eta^{*}} \\ \\
&\iff \bar{x} \geq \eta^{*}
\end{align}
So now we can define our rejection region in terms of the statistic $\bar{X}$:
$$R_{NP}=\left\{x:\bar{x} \geq \eta^{*} \right\}$$
where $\eta$ is chosen so that $P(\bar{X} \geq \eta^{*} \, | \theta_0)=\alpha.$
Now, under $H_0$ (or given $\theta_0=3$) we know that $\bar{X} \sim \mathcal{N}(3,1/n)$, to the above probability becomes:
\begin{align}
\alpha &= P(\bar{X} \geq \eta^{*} \, | \theta_0) \\
&= P \left(\frac{\bar{X}-3}{1\ \sqrt{n}} \geq \frac{\eta^{*}-3}{1\ \sqrt{n}}\right) \\
&= P \left(Z \geq \frac{\eta^{*}-3}{1\ \sqrt{n}}\right) \, \mbox{where } Z\sim \mathcal{N}(0,1)\\
\end{align}
Therefore, $\eta^{*} = \frac{z_{1-\alpha/2}}{\sqrt{n}}+3$, and we
$$\mbox{reject } H_0 \mbox{ iff } \bar{X} \geq \frac{z_{1-\alpha/2}}{\sqrt{n}}+3$$
So, in this case our rejection region depends on our test statistic ($\bar{X}$), the distribution associated to it (represented by $z_{1-\alpha/2}$, where $z$ is the critical value for a standard Normal), the sample size (n), and the value of our null hypothesis ($\theta=3$).
Example 2
Now, suppose we have the same setting from Example 1, but now $H_0: \theta = 4$ v.s. $H_1: \theta = 5$.
Using the same argument as above, we can see that $\eta^{*} = \frac{z_{1-\alpha/2}}{\sqrt{n}}+4$, and we
$$\mbox{reject } H_0 \mbox{ iff } \bar{X} \geq \frac{z_{1-\alpha/2}}{\sqrt{n}}+4$$
We changed the value of our null hypothesis, but everything else remained the same. The same way, our rejection region depends on our test statistic ($\bar{X}$), the distribution associated to it (represented by $z_{1-\alpha/2}$, where $z$ is the critical value for a standard Normal), the sample size (n), and the value of our null hypothesis ($\theta=4$).
Example 3
Suppose now that $X_1,\ldots,X_n \overset{iid}{\sim} \mbox{Exp}(\theta)$, and we wish to test the hypothesis $H_0: \theta = 3$ vs. $H_1:\theta=5$.
As pointed out by @Scortchi, this hypothesis is not the same as the previous one because on our first example we devised a test for the mean of our normal distribution, and now we are testing a value for the rate of our exponential distribution.
Using the same argument, one can show that we reject $H_0$ iff $\bar{X} \leq \eta^{*}$, where
$\eta^{*}$ is such that (under $H_0$)
$$P(\bar{X} < \eta^{*})=\alpha, \, \, \mbox{with} \,\, \bar{X} \sim \mbox{Gamma}(n,3n)$$
In this case our rejection region depends on our test statistic ($\bar{X}$), the distribution associated to it (Gamma, in this case), the sample size (n), and the value of our null hypothesis ($\theta=3$).
