Yes, that is correct. From the definition of what the Radon-Nikodym derivative is (which is essentially a type of anti-integral), for any event $\mathcal{A} \in \mathscr{X}$ and any measures $\pi$ and $\mu$ obeying the requirements you have given, we must have:
$$\begin{align}
\pi(\mathcal{A})
&= \int \limits_\mathcal{A} \frac{d\pi}{d\mu} \ d \mu \\[6pt]
&= \int \limits_\mathcal{A} \text{constant} \ d \mu \\[6pt]
&= \text{constant} \times \int \limits_\mathcal{A} \ d \mu \\[12pt]
&= \text{constant} \times \mu(\mathcal{A}), \\[6pt]
\end{align}$$
which shows that the measures are proportionate in this case, with the constant of proportionality equal to the constant value for the Radon-Nikodym derivative. If $\pi$ and $\mu$ are both probability measures then they must both obey the norming axiom of probability, which means they must both have a total measure of one. In this case you have $\text{constant} = 1$ and so $\pi = \mu$ (i.e., they are the same probability measure). (This is why when we work with probability, showing proportionality is sufficient to show that two distributions are the same.)