Let $\ \lambda(k)\ $ be the Carmichael function and denote $\ N(n)\ $ to be the number of solutions of $\ \lambda(k)=n\ $
Can we prove that $\ N(n)=n\ $ only holds for $\ n=100\ $ (besides the uninteresting case $\ n=0\ $) ?
Odd numbers can be ruled out since $\ N(1)=2\ $ and $\ N(u)=0\ $ for every odd $\ u\ge 3\ $.
A number $\ n\ $ with $\ n\equiv 2\mod 4\ $ cannot be a solution because of $\ N(2)=6\ $ and for $\ n\ge 6\ $, we can easily show that $\ N(n)\ $ must be divisible by $\ 4\ $. Hence , we can concentrate on the multiples of $\ 4\ $. Maybe someone finds additional restrictions.
Using recursive computation of $\ N(n)\ $ , I verified the conjecture upto $\ n=10^7\ $. Does anyone have an idea why $\ 100\ $ is so special here ?