TITLE: Equivalence of forcing automorphisms
QUESTION [5 upvotes]: Suppose that $P$ is a forcing poset in $V$. If $\pi$ is an automorphism of $P$ then $\pi$ extends to automorphisms of the names by induction: $$\pi\dot x = \lbrace(\pi p,\pi\dot y)\mid (p,\dot y)\in\dot x\rbrace$$
I've been stuck on the following proposition for quite some time now and I don't see an argument for, nor an obvious counterexample again:

Let $\mathscr G$ be a group of automorphisms of $P$. Suppose that $\dot x$ is a $P$-name and $G$ is $P$-generic over $V$ then the equivalence relation over $\mathscr G$ defined as: $$\pi\sim_G\sigma\iff (\pi\dot x)^G=(\sigma\dot x)^G$$ is multiplicative? Namely, is the set $\lbrace\pi\in\mathscr G\mid\pi\sim_G\mathrm{id}_P\rbrace$ is a normal subgroup of $\mathscr G$?

I am particularly interested in the case where $P$ is a Cohen forcing, in case it isn't true in general.
Edit: To restrict the question even more (after Joel's answer), what if we assume that $\dot x$ is hereditarily symmetric with respect to a normal filter of subgroups over $\mathscr G$?

REPLY [4 votes]: I think Joel's example can be easily modified to use a hereditarily symmetric name.  Use Cohen's original model for the negation of choice, i.e., use Cohen forcing to add an $\omega$-sequence of generic reals, and let the filter be the finite-support filter on the group $\mathcal G$ of permutations of the positions in the $\omega$-sequence.  Then use Joel's example working on  the first real in the $\omega$-sequence.  (If you don't like the fact that Joel's $\pi$ and $\tau$ aren't in $\mathcal G$, then adjoin them and extend the filter accordingly.)