If $\{w^k|w\in L\}$ regular implies L regular?

Question

If L is a language and the language $$\tilde{L}:=\{x^k,x\in L, k\in\mathbb{N}\}$$ is regular, does that imply that L is regular? ($|L|<\infty$ gives equivalence)

We came across this question when trying to prove an explicit example, namely

Prove that $L=\{a^{p^k}\mid p\text{ prime}, k\in\mathbb{N}\}$ is not regular.

It was fairly easy to show that $L_1=\{\underbrace{a^{p^1}}_{=a^p}\mid p\text{ prime}\}$ is not regular. If the above statement were true, this would give a very easy proof for the assignment.

I am asking out of interest, there are other ways to solve the question (I am NOT asking for answers on this example). I tried proving it via the pumping lemma, explicit construction of a FSA and the Myhill-Nerode theorem, however I was unable to find any notable results. I am unsure if this statement is true or if we can find a counterexample (I have been unable to do so)

Answer 1

No. Consider $L = \{a^{n^2} \colon n \ge 1\}$. As $a \in L$, your $\tilde{L} = \mathcal{L}(a^*)$, which is regular, but $L$ isn't.

Answer 2

Take any nonregular language $L$ of $a^*$ containing $a$. Then $\tilde L = a^*$ is always regular. This gives little hope for your problem.