Ah, that explains it, I think (no, see below) – the standard Nano (and its clones) have a low-dropout voltage regulator between VIN and 5V, but that still needs a margin of at least one volt, so if you pass in 5 V to VIN you get a lower voltage out (if you have 5 V, you’re supposed to connect it to 5V, not VIN). The processor still works, but odds are you’ll get something just below 4 V on the “5V” pin.
However, the newer Every design has a fancy switch-mode converter ($2.80 ea from Mouser) which apparently handles this better (it handles higher voltages a lot better). Cannot find any “drop-out equivalent” in the datasheet (all their examples use much higher supply voltages than the output voltages), but sounds like it’s close to zero for this case (or maybe not, see below). No, me read bad – it says max output is 80-85% of input, so with 5 V in you’d expect at least a 0.75 V drop. See below for more.
(but as long as everyone is using the same voltage, this should be neutral for the code – A0 and friends use the supply voltage as the reference voltage (= input value 1023) by default, and the voltage divider across the rotary switch splits up the same voltage, so the code will see the same values no matter the actual voltage on the “5V” pin. The important thing is that they’re evenly distributed.)