For ease of typing I’m going to remove the subscript $1$, since it plays no role in this question.
If you actually need $L'$ to be a distinct copy of $L$ and want to be somewhat rigorous about it, you’ll have to do some extra writing. You have (after minor modification) $L = (Q,E,\delta,q_0,F)$; I’m assuming that $Q$ is the state set, $E$ the input alphabet, $\delta:Q \times E \to Q$ the transition function, $q$ the initial state, and $F$ the set of acceptor (final) states. First you need a disjoint copy, $Q'$, of $Q$. A standard trick for constructing such a $Q'$ is to let $Q' = Q \times \{Q\} = \{\langle q,Q \rangle:q \in Q\}$; this ensures that $Q$ and $Q'$ are disjoint while at the same time providing a natural bijection between them. $Q'$ will be the state set of $L'$, and from that it’s easy to specify the rest of $L'$.
Certainly $L'$ will have the same input alphabet $E$. For convenience let $q' = \langle q,Q \rangle \in Q'$ for each $q \in Q$. Clearly we want $q_0'$ to be the initial state of $L'$ and $F' = \{q':q \in F\}$ to be the set of acceptor states. The transition function $\delta':Q' \times E \to Q'$ is defined by setting $\delta'(q',e) = \left(\delta(q,e)\right)'$. With these definitions $L' = (Q',E,\delta',q_0',F')$ is easily seen to be a DFA isomorphic to but disjoint from $L$, i.e., a distinct copy of $L$.
For your purpose you may or may not actually need this level of rigor in constructing $L'$. You would need it if, for instance, you wanted to combine two copies of $L$ to make some larger DFA.