Assume we have a (3-dimensional) resistor network that is mirror symmetric with respect to a plane $p$. The symmetry is only broken by a battery that creates a potential difference between points $A$ and $A'$ where $A'$ is the mirror image of $A$. Let $P$ be any point in the plane $p$. I would expect that for any point $Q$ on the network the potential difference between $Q$ and $P$ is identical to the potential difference between $P$ and $Q'$ (where $Q'$ is the mirror image of $Q$).
However, just mumbling "symmetry ..." does not really cut it for me. After all the symmetry is broken by the battery: there is definitely a net current from one side of the plane to the other.
Can you provide a general argument (or maybe some good intuition) that supports my expectation? Or disprove it?
And is it still true if there are multiple batteries (either connecting mirror images or occurring in symmetric pairs)?