Inspired By Garrett Clarke

In my previous solution to the Garrett's problem , I proved that for natural number $n$ with prime factorization $\prod_{i=1}^{k} p_i^{a_i}$ we have $n'=n \left(\frac{a_1}{p_1}+ \frac{a_2}{p_2}+... + \frac{a_k}{p_k} \right) \qquad(1)$ For this problem, note that $0$ is a solution, because all $p$ -th derivatives of $0$ is zero. Now, we need to have $n=n'$ , which from equation $(1)$ gives $\frac{a_1}{p_1}+ \frac{a_2}{p_2}+... + \frac{a_k}{p_k}=1 \qquad(2)$

We must have $p_k \ge a_k$ . Multiply both sides of equation $(2)$ by $p_1 p_2 ... p_{k-1}$ . It follows that $p_1 p_2 ... p_{k-1}\frac{a_k}{p_k}$ is an integer. Thus $p_k | a_k$ . Hence $a_k \ge p_k$ and $a_k=p_k$ . Subsequently $a_1=a_2=...=a_{k-1}=0$ and $n=p^p$ for some prime number $p$ .

Then its easy to see that answers to this problem are $0, 2^2, 3^3, 5^5, 7^7$ . Thanks @Abhinav Raichur for the beautiful problem.