Is the power of a regular language regular? Is the root of a regular language regular?

If $A$ is a regular set, then:

$L_1=\{x\mid\exists n \geq0, \exists y \in A: y=x^n\}$,

$L_2=\{x\mid \exists n \geq0, \exists y\in A: x=y^n\}$.

Which one of them is regular?

My reasoning is since in $L_2$ we can have uncountable $x$ from even one value of $y\ (y^0, y^1, y^2,...),\ L_2$ cannot be regular. But that thinking seems wrong.

The language $L_2$ is not necessarily regular. Indeed, consider $A = a^*b$. If $L_2$ were regular, then so would the following language be:
$$L_2 \cap a^ba^b = \{ a^nba^nb : n \geq 0 \}.$$
However, this language is not regular (exercise).

In contrast, the language $L_1$ is regular. We can see this by constructing a DFA for it. Let the DFA for $L_1$ have states $Q$, initial state $q_0$, accepting states $F$, and transition function $\delta$. The states of the new DFA are labeled by functions $Q \to Q$. The idea is that the new DFA is at state $f\colon Q \to Q$ if the original DFA moves from state $q \in Q$ to state $f(q)$ after reading $w$ (i.e., if $\delta(q,w) = f(q)$ for all $q \in Q$). The initial state is the identity function. When at state $f$ and reading $\sigma$, we move to the state $g$ given by $g(q) = \delta(f(q),\sigma)$. A state is accepting if $f^{(n)}(q_0) \in F$ for some $n \geq 0$.

StackExchange Computer Science Q#99341, answer score: 6