TITLE: Are generic filters that produce the same forcing extension related by a ground-model automorphism?
QUESTION [7 upvotes]: Suppose $M$ is a countable transitive model of some fragment of $\mathbf{ZFC}$, $\mathbb{P}\in M$ is a forcing notion and $G, H$ are $\mathbb{P}$-generic such that $M[G]=M[H]$. Does it then follow that there is some automorphism $\pi:\mathbb{P}\longrightarrow\mathbb{P}$ such that $\pi\in M$ and $\pi[G]=H$?
If the answer is no, are there natural restrictions one could impose on $\mathbb{P}$ (maybe apart from $\mathbb{P}$ being finite) such that the above sentence holds?

REPLY [12 votes]: This works much better in terms of complete Boolean algebras. If $\mathbb B$ and $\mathbb B'$ are complete Boolean algebras and a $\mathbb B$-generic filter $G$ and a $\mathbb B'$-generic filter $G'$ generate the same forcing extension, then there are $b\in G$ and $b'\in G'$ such that the part of $\mathbb B$ below $b$ is isomorphic to the part of $\mathbb B'$ below $b'$ by an isomorphism in the ground model, that sends the restriction of $G$ to the restriction of $G'$.
I believe this fact is in Serge Grigorieff's paper "Intermediate submodels and generic extensions in set theory" [Ann. Math. Second Series, Vol. 101, No. 3 (May, 1975), pp. 447-490].
It's certainly possible to rewrite this in terms of posets instead of complete Boolean algebras, but the result looks messy to me.