A graph $G$ with two $K_6$ subgraphs, in which any one-factor of $G$ induces a one-factor in exactly one of the $K_6$ subgraphs? I'm seeking a simple graph $G$ of the following type:


*

*It contains two disjoint copies of $K_6$ (the complete graph on 6 nodes), $H$ and $H'$ say.

*Any one-factor of $G$ must contain either (a) a one factor of $H$ and no edges in $H'$ or (b) a one factor of $H'$ and no edges in $H$.

*There exists a one-factor of $G$ that contains a one-factor of $H$.

*There exists a one-factor of $G$ that contains a one-factor of $H'$.



Question: Does $G$ exist?

I'm also interested in the same problem with $K_6$ replaced by $K_{2n}$.
My motivation for this question comes from an attempt to rephrase a question about Latin squares as a question about one-factors.
 A: Here is a proof that $G$ does not exist.  Let $M_1$ be a (red) 1-factor which contains a 1-factor $M_1'$ of the first $K_6$, and $M_2$ be a (blue) 1-factor which contains a 1-factor $M_2'$ of the second $K_6$.  Note that $M_1 \triangle M_2$ is a union of even cycles, with alternating red and blue edges.  Observe that we can get a new 1-factor by 'flipping' the edges on any such cycle.  Therefore, there must be a cycle $C$ of $M_1 \triangle M_2$ which contains all the edges of $M_1' \cup M_2'$, otherwise we violate the second condition.  Let $e \in E(C) \cap E(M_1')$.  Traverse $C$ clockwise and let $f$ be the next edge of $C$ which is also in $E(M_1')$.  Let $P$ be the subpath of $C$ beginning at $e$ and ending at $f$.  Note that there is an edge $g$ between the ends of $P$ because of the first $K_6$.  Therefore, if we flip the edges of $M_1$ along the even cycle $P \cup g$, we get another 1-factor.  However, the intersection of this 1-factor with the first $K_6$ is neither empty nor a 1-factor.  
Note that this proof works for any $K_{2n}$ for $n \geq 2$.  
