$\underline{\text{Is it true that the entropy of a measure is equal to the entropy of its push-forward by some measurable map?}}$
Without all the details, let $\mu$ be some probability measure on $\mathbb{R}^d$ (for simplicity let it be a density), $f:\mathbb{R}^d\to\mathbb{R}^d $ some measurable map, and $f_{\#}\mu$ the push-forward of $\mu$ by $f$ :https://en.wikipedia.org/wiki/Pushforward_measure . The Entropy of $\mu$ is defined as
$$ H(\mu)=\int \mu \log\mu ~dx. $$
Is it true $H(\mu)=H(f_{\#}\mu)$?
I thought so since : $f_{\#}\mu(x)=\mu \circ f^{-1}(x)$, moreover the change of variables gives
\begin{align*} H(f_{\#}\mu)=\int f_{\#}\mu (x)\log f_{\#}\mu(x) ~dx=&\int \mu(x) \log f_{\#}\mu \circ f(x) ~dx \\ =&\int \mu(x)\log \mu(x) dx. \end{align*}
Not true in general. For example, if $f$ is the constant $0$ map, the density of the pushforward is not defined (but, in a limiting sense, the entropy of the pushforward is $-\infty$). Even if $f$ preserves Lebesgue measure, it's not necessarily true. Here's an example of what can go wrong: Take $f$ to be the non-injective function shown in https://math.stackexchange.com/a/1855927/222048, and concentrate $\mu$ on a single one of the three branches. The density of $\mu$ gets stretched out, so the entropy increases.
However, if $f$ preserves Lebesgue measure in both directions then I believe the entropy is preserved. For example, in Hamiltonian dynamics, the time flow preserves volume in position-momentum phase space (Liouville's Theorem), so the Gibbs entropy remains constant. More generally, I believe this holds whenever the integrand depends on $x$ only via the density.