Prove EXP to be regular or non-regular

Question

Given L is regular, Prove/Disprove that the following language is regular or not. $EXP = \{w| w^{|w|} ∈L\}$

Answer 1

Suppose that $L$ is accepted by a DFA with states $Q$, initial state $q_0$, accepting states $F$, and transition function $\delta$.

With every word $w$ we can associate the function $\delta_w\colon Q \to Q$ given by $\delta_w(q) = \delta(q,w)$. Note that there are finitely many such function. It is not hard to check that for any such function $\delta_w$, the language $L_{\delta_w} = \{ x : \delta_x = \delta_w \}$ is regular.

We can also associate with $w$ a set $N_w = \{ n \in \mathbb{N} : \delta(q_0,w^n) \in F \}$. Note that $N_w$ depends only on $\delta_w$, and so there are finitely many such sets. Moreover, each of them is eventually periodic (why?), and so the language $L_{N_w} = \{ x : |x| \in N_w \}$ is regular.

Combining the two observations, we see that $$ EXP = \bigcup_{w \in \Sigma^*} (L_{\delta_w} \cap L_{N_w}). $$ While the sum is infinite, there are only finitely many different summands, and we deduce that $EXP$ is regular.