Consider the Cohen forcing, and suppose that $\dot x,\dot y$ are names for reals, which are not in the ground model (i.e. $1$ forces that neither is in the ground model).


Can we always find an automorphism mapping $\dot x$ to $\dot y$?


The answer is negative, as Andreas Blass points out. But let me refine the question a lot more.


Suppose $p$ is a condition which forces that $\dot x$ and $\dot y$ are both generic with respect to the restriction of the forcing to $A$ (i.e. take only conditions whose domain is a subset of $A$ and complete that to the subalgebra of the Cohen forcing). Is there an automorphism $\pi$ such that $p\Vdash\pi\dot x=^\*\dot y$?


Of course, this depends on how you consider the Cohen forcing, so let's take the "richest" way, and consider it as the completion of functions from finite sets of integers to $\{0,1\}$.



Can we always permute Cohen reals?
No, because reals not in the ground model can be different in definable (with parameters from the ground model) ways. For example, $\dot x$ might be (forced by all conditions to be) a Cohen-generic subset of $\omega$, while $\dot y$ is the intersection of $\dot x$ with the set of even numbers. Then $\dot y$ is disjoint from an infinite ground-model set of natural numbers but $\dot x$ is not.


EDIT to answer the edited version of the question: The answer is still negative, if $A$ is an infinite, coinfinite, ground-model subset of $\omega$. Let $\dot c$ be (the canonical name of) the Cohen subset of $\omega$ directly added by the forcing. Let $\dot x$ be the intersection of $\dot c$ with $A$. Let $\dot y$ be a copy on $A$ of all of $\dot c$, i.e., the $n$-th element of $A$ is in $\dot y$ iff $n\in\dot c$. Then both $\dot x$ and $\dot y$ are generic with respect to the Cohen forcing over $A$. They are not related by an automorphism of the forcing, because $\dot y$ generates (over the original ground model) the whole forcing extension (the same as $\dot c$) while $\dot x$ generates only an intermediate model.


