Let $L$ be a regular language, and let
$$
L_1 = \{u_1\#u_2\# \dotsm \#u_n \mid ∃v_1,v_2,…,v_{n-1} \in \Sigma^3 \text{ such that } u_1v_1u_2 \dotsm v_{n-1}u_n \in L \}
$$
where $\# \notin \Sigma$.
For example: $\Sigma = \{0,1\}$, $L = \{010101\}$, $L_1 = \{\#\#, 01\#1, 0\#01\}$.
My thought of the process was to try to validate by this logic. Let's say I get the word $z = 01\#1$. Then I check how many characters I passed before reaching $\#$, if I passed $n \leqslant 2$ then $u_1 = 01$, and I send it to $L$ automata for validation. Then I skip $\#$ and continue with the same logic. I'm not sure my solution is working, and I can't think of a way to draw it.
The following construction should work. Start with a finite automaton $\cal A$ for $L$ and, for each path $p_0 \xrightarrow{a_1} p_1 \xrightarrow{a_2} p_2 \xrightarrow{a_3} p_3$ in $\cal A$, add a new transition $p_0 \xrightarrow{\#} p_3$. The resulting non deterministic automaton recognises $L_1$.