What happens when salt water is electrolyzed with a copper anode and a copper cathode?
The chemistry here can be explained from the associated products of the electrolysis of brine. To quote an education source:

Three important chemicals, NaOH, Cl2, H2, can be obtained by electrolyzing an aqueous NaCl solution (brine). This forms the basis of the chlor-alkali industry. The diaphragm cell (also called a Hooker cell)...

As such, depending on the cell design, expect also hypochlorous acid:
$\ce{Cl2 (g) + H2O (l) &lt;-&gt; HCl (aq) + HOCl(aq) }$
and hypochlorite with OH- as a side product of the electrolysis of the brine:
$\ce{ OH- + HOCl -&gt; H2O + ClO-}$
However, some of the $\ce{HOCl}$ can also engage in a Fenton-type reaction (example here) with $\ce{Cu}$ (and, to a lesser extent, $\ce{HOCl}$ acting on $\ce{Cu2O}$ resulting in some possible $\ce{CuO}$ or even a basic chloride formation) generally depicted as follows:
$\ce{Cu/Cu+ + HOCl -&gt; Cu+/Cu++ + .OHCl-}$
where the latter radical intermediate can further break down with products dependent on the pH level (here, likely over pH 7 from $\ce{OH-}$ created in the parallel electrolysis reaction) and the chloride concentration. Source: See reactions listed in Supplement Table S1 available here by clicking on Supplement File to obtain a download.
Support: With metal spoons serving as electrodes for the electrolysis of a saltwater solution, there is clear discoloration of one of the spoon electrodes (visible at 5 minutes and 20 seconds into this science experiment video).
Bottomline in time, expect similar discoloration of a copper electrode as well from the associated electrolysis products.