Consider the following scenario: one has 2 communication channels $C\_1$ and $C\_2$. Denote by $p(x)$ the input probability distribution.


The mutual information between the input and the output of $C\_1$ must be greater or equal than the mutual information between the input and the output of the composed channel $C\_1\circ C\_2$ (i.e. act with $C\_1$ first then feed the output to $C\_2$). This follows from the data processing inequality. 


My questions are:


1. Given $p(x)$, what are the channels $C\_2$ for which equality holds (i.e. mutual information is non-decreasing). I know from of proof of data processing inequality that is true if and only if one has a Markov chain, but what can we say about the conditional transition matrices, i.e. relations between $p(x)$, $C\_1$ and $C\_2$? Or, in other words, what is the functional form of $C\_2$ as a function of $C\_1$ and $p(x)$?
2. And the reverse: given $C\_2$, what are the input distributions $p(x)$ for which the mutual information is non-decreasing?


I wasn't able to find an elegant solution to this problem, I have only some partial solutions. For example, if $C\_2$ is a permutation channel, then mutual information stays the same no matter what $p(x)$ is. Thanks!


PS: I hope it is clear what I mean by mutual information between the input and output of a channel, it is the mutual information of the joint probability distribution obtained by multiplying the elements of the transition matrix with the corresponding component of the input, $p(x,y)=p(y|x)p\_{0}(x)$. 



Mutual information staying constant under composition of channels
I only know the answer for the first one.
If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.


Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this post[1](https://mathoverflow.net/questions/177312/double-markovity))


