How to solve $x^x = y^y \mod p$?
question 1: $x\equiv y$ is the trivial answer. knowing parity arguments seems more useful, odd mod odd is opposite the integer multiplier for example. So $3^3$ is 6 mod 7 so it's multiplier is odd (3 in fact). $3^5$ is 5 mod 7 so it's multiplier is even (34 which is 6 mod 7 in this case). So odd pairs (x,y) depend on being in the same parity multiplier. You can do similar mod 3 etc. That will narrow it down.