2023 AMC 12A Problem 14

Taking magnitudes gives $|z|^5=|z|,$ so $|z|=0$ or $|z|=1.$ The value $z=0$ works, giving one solution. If $|z|=1,$ multiply the equation by $z$ to get $z^6=z\overline{z}=|z|^2=1.$ This has $6$ distinct roots, all of modulus $1.$ Altogether there are $1+6=7$ solutions. Thus, the correct answer is E.