Surprisingly (to me), this problem seems to have a simple solution via graph theory.

Imagine we start with a logistic game $p$, and we want to modify it to get another logistic game $q$ in which each player retains the same overall probability of winning. There must exist at least one pair of players $i, j$ such that we assign $i$ a higher probability of beating $j$ in the $q$ than in $p$.

Since there is a player that has been made stronger against $j$, in order for $j$'s overall win probability to remain the same, $j$ must be made stronger against some other player. That player must then be made stronger against some other player, and so on. In this way we can walk through the finite set of players. Since the set is finite, at some point we must visit a previously visited player. But this implies a loop in which $i$ is stronger against $j$, $j$ is stronger against $k$, ...is stronger against $i$. The existence of this kind of loop violates odds transitivity.

Surprisingly (to me), this problem seems to have a simple solution via graph theory.

Imagine we start with a logistic game $p$, and we want to modify it to get another logistic game $q$ in which each player retains the same overall probability of winning. There must exist at least one pair of players $i, j$ such that $q(i, j) > p(i, j)$.

Since there is a player that has been made stronger against $j$, in order for $j$'s overall win probability to remain the same, $j$ must be made stronger against some other player. That player must then be made stronger against some other player, and so on. In this way we can walk through the finite set of players. Since the set is finite, at some point we must visit a previously visited player. But this implies a loop in which $i$ is stronger against $j$, $j$ is stronger against $k$, ...is stronger against $i$. The existence of this kind of loop violates odds transitivity.