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.

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 $\mathscr G$?

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 the set $\lbrace\pi\in\mathscr G\mid\pi\sim_G\mathrm{id}_P\rbrace$ is a normal subgroup.

I am particularly interested in the case where $P$ is a Cohen forcing, in case it isn't true in general.

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.