Suppose that $G=MN$ and $G=MP$ are two exact factorization of a finite group $G$. What is the relation between $M$ and $P$? Clearly if $G=MN$ then $G=M(mNm^{1})$ is another factorization of $G$. Is this the only possibility to change $N$ into $P$?

The answer of the problem is the following: A group G has two exact factorizations $G = M N = M P$ if and only if there exists a unitary bijective map (just a map, not a morphism of groups) $v : N \to P$ such that $n v(n)^{1} \in M$, for all $n \in N$ and $v$ satisfy four natural compatibility conditions (not very transparent to write down here  they are similar, mutatis mutandis, to the one given in Propositions 2.1 in http://front.math.ucdavis.edu/0903.5060  for example one of them is that v is a morphism of $M$sets). 


The answer is no. $G=MN$ is an exact factorisation is equivalent to $N$ acting regularly on the set of right cosets of $M$ in $G$. It is not necessary for two regular subgroups of a group to be isomorphic. Think for example $G=S_n$ and $M=S_{n1}$. Then any group of order $n$ acts regularly on $n$ points so provides a group for $N$. 


Does "exact factorization" mean every element of $G$ has exactly one expression as $mn$ with $m$ in $M$ and $n$ in $N$? If not, ignore the rest of this answer. But if so, then let $G$ be the group of the square, let $M$ be the subgroup of rotations, and let $N$ be any one of the four subgroups generated by a flip. Then $G=MN$ but the four available $N$ are not all conjugate to each other. 


I am not an expert in group theory. The problem is interesting and is a particular case to the following, which is a little bit more general: when two bicrossed products of two groups are isomorphic? (a group G has an exact factorization iff it is a bicrossed product of two (sub)groups  see a Tackeuchi's old theorem). A partial answer to the problem (in fact a Schreier type theorem for the factorization/bicrossed product) was given in http://front.math.ucdavis.edu/0903.5060 In this elementary paper the problem is solved in the case that the isomorphism fix (stabilize) one of the factors. PS (edit): I think we can get a full answer of your question if we apply Corollary 3.11 and Theorem 3.3 from the above paper (writen equivalent in the language of bicrossed products, sorry for the categorical stuff used there). The identity is an isomorphism of the group $G$ that fix (satabilize) M. Thus we can apply for it Corollary 3.11. Cheers! Gigel Militaru 


What a regular action means?, transitive withous fixed points? And what a regular subgroup is? If $M$ is normal in $G$ then both $N$ and $P$ are isomorphic to $G/M$. 

