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 $\dot y$?

If the answer is negative, as I suspect it is, is there some reasonable condition on the names in order to have the answer yes?

Of course, this depends on how you consider the Cohen forcing, so let's take the "richest" way, and consider it as the countable atomless Boolean algebra.

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 $\{0,1\}$.

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 $\dot y$?

Of course, this depends on how you consider the Cohen forcing, so let's take the "richest" way, and consider it as the countable atomless Boolean algebra.

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 $\dot y$?

If the answer is negative, as I suspect it is, is there some reasonable condition on the names in order to have the answer yes?

Of course, this depends on how you consider the Cohen forcing, so let's take the "richest" way, and consider it as the countable atomless Boolean algebra.