Is entropy conserved under invertible mappings? Suppose for a random variable $ X\colon \Omega \to E $, I have an invertible mapping $ Z = f(X) $.
Is the Shannon Entropy for each variable equivalent?
$$ H(X) = H(Z) $$
If not, can anything relationship between the entropies be said in general (eg. proportionality, additive constant, etc.)?
 A: 
Is the Shannon Entropy for each variable equivalent?

Yes. An intuitive way of seeing why this is so is that (deterministic) transformations cannot increase entropy. 
Specifically, note that $$H(X) \geq H(f(X)) = H(Z),$$ because, by the chain rule of entropy:
$$
H(X, f(X)) = H(X) + H(f(X) | X) = H(X),
$$
but also
$$
H(X, f(X)) = H(f(X)) + H(X | f(X)) \geq H(f(X))
.
$$
However, letting $g = f^{-1}$ (which exists, as $f$ is invertible), also $$H(Z) \geq H(g(X)) = H(X).$$
A: The answer given by Ami is correct, I will provide an intuitive explanation of why $H(X|Z) = H(Z|X) = 0$: 


*

*$f(\cdot)$ is a deterministic mapping from $X$ to $Z$, so when you observe $X$, you have zero uncertainty to infer about $Z$, thus $H(Z|X) = 0$.

*Note that $Z=f(X)$ is invertible, so when you observe $Z$, you also have zero uncertainty to infer about $X$, that's to say, $H(X|Z)=0$.


Combining 1 and 2, it is clear that $H(X)=H(X,Z)-H(Z|X)=H(X,Z)$ and $H(Z)=H(X,Z)-H(X|Z)=H(X,Z)$, so $H(X)$ and $H(Z)$ are exactly the same.
