Is it true that for every regular language $L \subseteq \{0, 1\}^{*}$ the language $\{w^{|w|} |w \in L \}$ is also regular?
I think the language is not regular because for a finite w we can easily construct DFA by chaining |w| DFAs that accept w but |w| is not bounded therefore we need a non-finite automaton !
In fact, it is not even true that for every regular language $L \subseteq \{1\}^*$, the language $L' = \{w^{|w|} \mid w \in L\}$ is regular.
In particular, let $S_V = \{n \in \mathbb{N} \mid 1^n \in V\}$ for all languages $V \subseteq \{1\}^*$. Note that $V = \{1^n \mid n \in S_V\} = \{s \in \{1\}^* \mid |s| \in S_v\}$.
Clearly, we see that $S_{L'} = \{n^2 \mid n \in S_L\}$ for all $L \subseteq \{1\}^*$.
Now take $L = \{1\}^*$. Then we see that $S_{L'} = \{n^2 \mid n \in \mathbb{N}\}$. That is, $L'$ is the set of all strings whose length is a perfect square.
I will leave it as an exercise to apply the pumping lemma and finish the proof.