This is too long for a comment, sorry.
To summarize, you have RVs $U-X-Y-V$ where:
- $X\sim B(1/2)$,
- $Y=X+Z_Y$
- $U=X+(X\cdot Z_{U1} + (1-X)\cdot Z_{U0})$,
- $V=Y+(Y\cdot Z_{V1} + (1-Y)\cdot Z_{V0})$,
for RVs:
- $Z_Y\sim B(p_Y)$,
- $Z_{Ui}\sim B(p_{Ui})$, $i=0,1$
- $Z_{Vi}\sim B(p_{Vi})$, $i=0,1$
You are essentially wondering whether $U',V'$ with $U'-X-Y-V'$ and:
- $U'=X+Z_{U'}$
- $V'=Y+Z_{V'}$
choosing:
- $Z_{U'}\sim B(p_{U'})$ so that $I(U';X)=I(U;X)$
- $Z_{V'}\sim B(p_{V'})$ so that $I(V';Y)=I(V;Y)$
always gives $I(U';V')\geq I(U;V)$.
If all the $Z$'s turn out to be independent because of the Markov structure (not sure whether or not this is true), my inclination is that the answer to your question is always no for nontrivial cases, i.e. when (U,X) or (Y,V) are not BSCs
This is because there is a notion that uniform noise is the most degrading for any given link, so choosing $(X,U'),(Y,V')$ to be BSCs will make the noise distribution $V'-U'$ uniform.