Let $n$ have that property. Then $a \equiv a \mod n$ gives $a^2 \equiv 1 \mod n$ for all $a$ coprime to $n$. Since $2^2 \not\equiv 1 \mod 5$ (and because we always replace $2$ by some residue coprime to $n$ and $\equiv 2 \mod 5$) we get $5 \nmid n$. Especially we are forced to have $5^2 \equiv 1 \mod n$, thus $n|24$. It's trivial that $n=3$ and $n=8$ work, thus by the chinese remainder theorem the same holds for $n=24$. This imediately implies the property for the divisors of $24$, too.

you can easily find that if $ a\leq{\sqrt {n}}$ then $ \gcd(a,n) > 1$. It means that all primes $ \leq{\sqrt {n}}$ divides n. but it's known that if $ \sqrt {n} > 10$ then there exist at least two distinct primes $ p$ and $ q$ such that $ \frac {\sqrt {n}}{2} < p < q < \sqrt{n}$ but these primes must divide $ n$ it follows that $ n$ must be one of these forms: $ 3pq,2pq$ (because $ \frac {n}{4} < pq < n$ and $ n = kpq$ so...). the rest is trivial.