Call a number good if it is not bad. We claim all good numbers are products of distinct primes, none of which are equivalent to 1 modulo another. We first show that all such numbers are good. Consider $n=p_{1} p_{2} \ldots p_{k}$, and let $x$ be a number satisfying $x \equiv c\left(\bmod p_{1} p_{2} \ldots p_{k}\right)$ and $x \equiv 1\left(\bmod \left(p_{1}-1\right)\left(p_{2}-1\right) \ldots\left(p_{k}-1\right)\right)$. Since, by assumption, $p_{1} p_{2} \ldots p_{k}$ and $\left(p_{1}-1\right)\left(p_{2}-1\right) \ldots\left(p_{k}-1\right)$ are relatively prime, such an $x$ must exist by CRT. Then $x^{x} \equiv c^{1}=c$ $(\bmod n)$, for any $c$, as desired. We now show that all other numbers are bad. Suppose that there exist some $p_{1}, p_{2} \mid n$ such that \operatorname{gcd}\left(p_{1}, p_{2}-1\right) \neq 1$(which must hold for some two primes by assumption), and hence \operatorname{gcd}\left(p_{1}, p_{2}-1\right)=p_{1}$. Consider some $c$ for which $p_{1} c$ is not a $p_{1}$ th power modulo $p_{2}$, which must exist as $p_{1} c$ can take any value modulo $p_{2}$ (as $p_{1}, p_{2}$ are relatively prime). We then claim that $x^{x} \equiv p_{1} c(\bmod n)$ is not solvable. Since $p_{1} p_{2} \mid n$, we have $x^{x} \equiv p_{1} c\left(\bmod p_{1} p_{2}\right)$, hence $p_{1} \mid x$. But then $x^{x} \equiv p_{1} c$ is a $p_{1}$ th power modulo $p_{2}$ as $p_{1} \mid x$, contradicting our choice of $c$. As a result, all such numbers are bad. Finally, it is easy to see that $n$ is bad if it is not squarefree. If $p_{1}$ divides $n$ twice, then letting $c=p_{1}$ makes the given equivalence unsolvable. Hence, there are 16 numbers ( 13 primes: $2,3,5,7,11,13,17,19,23,29,31,37$, 41 ; and 3 semiprimes: $3 \cdot 5=15,3 \cdot 11=33,5 \cdot 7=35)$ that are good, which means that $41-16=25$ numbers are bad.