Random mapping and entropy ordering Let $X_1,X_2$ be discrete random variables such that $H(X_1)<H(X_2)$ where $H()$ is the entropy. We know that for any random mapping $T$ which is invertible ($T$ is a function of $\omega$ and $X$, by invertible we mean invertible for every fixed $\omega$) and independent of a random variable $X$ the following is true $$H(T(X))\geq H(T(X)|T)=H(T^{-1}T(X)|T)=H(X),$$ and we have $$H(T(X_1))\geq H(X_1)\text{ and }H(T(X_2))\geq H(X_2).$$ If $X_1$, $X_2$, $T(X_1 )$ and $T(X_2 )$ have the same support, is it true that $$H(T(X_1))\leq H(T(X_2))? $$
The reason for assuming the same support is the following: If the cardinality of the support of $X_2$ is greater than the cardinality of the support of $X_1$ and  $P(T(X_1))$, $P(T(X_2))$ are uniform then the above does not hold. 
Please correct me if I made any mistakes. Thank you for your time.
 A: The following constructs a counterexample on three letters.
Take an $X$ that is distributed on $\{1,2,3\}$ under the law $(p_1, p_2, p_3)$, and let $T$ be picked uniformly at random from $\{T_a, T_b\}$ independently of $X$ for $T_a = (1,2,3), T_b = (2,1,3),$ where the vectors should be read as $(T_*(1), T_*(2), T_*(3))$. Notice that $T$ is a supported on invertible functions as desired. Further, the action of $T$ is to smooth out the mass on $\{1,2\}$ and to leave the mass on $3$ unchanged. Concretely, $$ P(T(X) = 1) = P(T(X) = 2) = \frac{p_1 + p_2}{2}, P(T(X) = 3) = p_3.$$
Now suppose we took an $X_1 \sim p$ with $p_3 = 1/3$ but $p_1$ and $p_2$ very skewed, and we took an $X_2 \sim q$ with $q_3 > 1/3$ but $q_1 = q_2$. Then $T(X_1)$ will be uniform but $T(X_2)$ has the same law as $X_2$. So as long as $H(X_2) > H(X_1)$ this yields a counterexample.
Concretely, take $X_1 \sim (p, 2/3 - p, 1/3)$ and $X_2 \sim (1/4, 1/4,1/2)$. By the argument above, $H(T(X_2)) = H(X_2) < \log_2(3) = H(T(X_1))$. So all we have to ensure is the existence of $\alpha$ such that $H(\alpha, 2/3 - \alpha, 1/3) < H(1/4, 1/4, 1/2)$. This exists by a continuity argument near $p = 0$, but we can be very explicit - take $\alpha = 1/6$. Then $H(X_1) = 2/3 + \log_2(3)/2 < 3/2 = H(X_2).$
