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It depends on the forcing notion, on the generic filter, on the group of automorphisms and on the name $\dot x$.

First, there are some trivial cases where it is normal, such as the case of a check name $\check x$, which is fixed by every automorphism $\pi$, and so that subgroup is the whole group and hence normal. Similarly, if the forcing notion $P$ is rigid, then it has no nontrivial automorphisms at all, in which case the subgroup again is normal.

Meanwhile, in the case of Cohen forcing, your main case, it is sometimes not normal, depending on the name $\dot x$. Consider the case of the canonical name for the generic object $\dot x=\dot G$, and let $\pi$ be an automorphism that acts only below a condition that happens not to be in the filter $G$. For example, perhaps $\pi$ swaps the second bit of the Cohen real, but only if the first bit is $0$, whereas the first bit of $G$ happens to be $1$. Thus, $(\pi\dot G)^G=(\dot G)^G=G$, since this operation has no effect on the part of the name relevant for $G$. Meanwhile, if let $\tau$ is be an automorphism that moves that cone into $G$, such as flipping the first bit of the Cohen real. Now Observe now that $(\tau^{-1}\pi\tau\dot G)^G\neq G$, since the former will flip the second bit of $G$, and so the subgroup is not normal.

A similar argument will work in many other cases, with highly homogeneous forcing, and so I think we should think of the property usually as failing except in very special circumstances. The way I think about it is this: your equivalence relation is able to ignore huge parts of the automorphism, since it is interpreting the name by the filter $G$, and $G$ is very local to particular conditions in the poset, but the normality requirement on the automorphisms is much more global, and so they will sometimes conflict.

1

It depends on the forcing notion, on the generic filter, on the group of automorphisms and on the name $\dot x$.

First, there are some trivial cases where it is normal, such as the case of a check name $\check x$, which is fixed by every automorphism $\pi$, and so that subgroup is the whole group and hence normal. Similarly, if the forcing notion $P$ is rigid, then it has no nontrivial automorphisms at all, in which case the subgroup again is normal.

Meanwhile, in the case of Cohen forcing, your main case, it is sometimes not normal, depending on the name $\dot x$. Consider the case of the canonical name for the generic object $\dot x=\dot G$, and let $\pi$ be an automorphism that acts only below a condition that happens not to be in the filter $G$. For example, perhaps $\pi$ swaps the second bit of the Cohen real, but only if the first bit is $0$, whereas the first bit of $G$ happens to be $1$. Thus, $(\pi\dot G)^G=(\dot G)^G=G$, since this operation has no effect on the part of the name relevant for $G$. Meanwhile, if $\tau$ is an automorphism that moves that cone into $G$, such as flipping the first bit of the Cohen real. Now $(\tau^{-1}\pi\tau\dot G)^G\neq G$, since the former will flip the second bit of $G$, and so the subgroup is not normal.

A similar argument will work in many other cases, with highly homogeneous forcing, and so I think we should think of the property usually as failing except in very special circumstances. The way I think about it is this: your equivalence relation is able to ignore huge parts of the automorphism, since it is interpreting the name by the filter $G$, and $G$ is very local to particular conditions in the poset, but the normality requirement on the automorphisms is much more global, and so they will sometimes conflict.