Under what conditions does the function C = f(A,B) satisfy H(C|A) = H(B)?
Note 

\begin{align}
0&amp;=H(C|A,B)\\
&amp;=H(A,B,C)-H(A,B)\\
&amp;=H(B|A,C)+H(C|A)+H(A)-H(A,B)\quad\text{(chain rule)}\\
&amp;=H(B|A,C)+H(C|A)-H(B|A),
\end{align}

so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)\ge 0$ and $H(B)\ge H(B|A)$, your condition is equivalently $H(B|A,C)=0\wedge H(B)=H(B|A)$. 

For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $\{b\mid \mathrm{Pr}\{A=a, B=b\}&gt;0\}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.