In general is $\frac{d\,\mu_1}{d\,\mu_2}\circ T = \frac{d\,T\mu_1}{d\,T\mu_2}$? Given an ergodic and non-singular dynamic system (definition provided here) $(X, \mathcal{B}, \mu_1, T)$ where $(X, \mathcal{B}, \mu_1)$ is a measure space and $T$ is a fixed transformation, we then will have $\mu_1$ equivalent to $T\mu$ where $T\mu_1$ is defined by $T\mu_1(A)=\mu_1[T^{-1}(A)]$ for each $A\in \mathcal{B}$. Now suppose $\mu_2$ is another measure that is equivalent to $\mu_1$ and makes $(X, \mathcal{B}, \mu_2, T)$ ergodic and non-singular. Hence, we will have $T\mu_2\sim\mu_2\sim\mu_1\sim T\mu_1$. Do we have:
$$
\frac{d\,\mu_1}{d\,\mu_2}\circ T = \frac{d\,T\mu_1}{d\,T\mu_2}
$$
$\mu_1$-almost everywhere? Here the "$\mu_1$-almost everywhere" can be replaced by "$\mu_2$-almost everywhere" or any one of those four measures. By definition of $T\mu_2$, for each $A\in\mathcal{B}$, we have:
$$
\int_X\chi_A(x)d\,T\mu_2 = \int_X\chi_A(Tx)d\,\mu_2
$$
I tried to show the following equation holds for each $A\in\mathcal{B}$ but cannot:
$$
\int_A\frac{d\,\mu_1}{d\,\mu_2}(x)d\,T\mu_2 = \int_A\frac{d\,\mu_1}{d\,\mu_2}\circ T(x)d\,\mu_2 = \int_A\frac{d\,T\mu_1}{d\,T\mu_2}(x)d\,\mu_2
$$
Any hints or counter-examples will be appreciated
 A: I am fairly sure this is not the case (unless I am missing something in the definition of $T$)
Consider the special case where $T$ is invertible with inverse $T_{\rm inv}$. Then, it holds for any test function $f$
\begin{aligned}
\int f \,dT\mu_1 &= \int f \circ T \,d\mu_1  \\
&= \int f \circ T \,\frac{d\mu_1}{d\mu_2} \,d\mu_2 \\
&= \int f \circ T \, \frac{d\mu_1}{d\mu_2} \,d(T_{\rm inv} (T \mu_2))\\
&= \int f \,\frac{d\mu_1}{d\mu_2} \circ T_{\rm inv} \,dT\mu_2
\end{aligned}
and thus it actually holds
$$
\frac{dT\mu_1}{dT\mu_2} = \frac{d\mu_1}{d\mu_2} \circ T_{\rm inv}.
$$
If I understand correctly, the following simple example satisfies your assumptions and shows that your claim does not hold: Take $X = \{1, 2, 3\}$, $\mu_1 = (1/6, 2/6, 3/6)$ (meaning $\mu_1 = \frac{1}{6}\delta_1 + \frac{2}{6} \delta_2 + \frac{3}{6} \delta_3$), $\mu_2 = (2/6, 3/6, 1/6)$ and $T(1) = 2, T(2) = 3, T(3) = 1$, and one quickly finds that $T$ is invertible but also that $\frac{dT\mu_1}{dT\mu_2} = \frac{d\mu_1}{d\mu_2} \circ T_{\rm inv} \neq \frac{d\mu_1}{d\mu_2} \circ T$.
