Suppose we have the following language operation:
$Duplicate(L) = \{Duplicate(w)|w \in L\}$ where if $w=w_1w_2\ldots w_n$ $Duplicate(w) = w_1w_1w_2w_2 \ldots w_nw_n$.
It is simple to construct a DFA for this language, but is it valid to solve it by simply renaming $\Sigma$? (I.e., if $\Sigma$ is the old alphabet in the original DFA construct a new DFA that is the same as the original DFA except $\Sigma' = \{\sigma\sigma|\sigma \in \Sigma\}$.)
What you are describing is not a change of alphabet but rather an inverse homomorphism. Define the homomorphism $h\colon \Sigma \to \Sigma^*$ by $h(\sigma)=\sigma^2$. Then $h(\operatorname{Duplicate}(L)) = L$ and also $h^{-1}(L) = \operatorname{Duplicate}(L)$; here $h(\operatorname{Duplicate}(L))$ is the result of applying $h$ to all words in $\operatorname{Duplicate}(L)$, and $h^{-1}(L)$ is the set of words $x$ such that $h(x) \in L$. Since the regular languages are closed under homomorphism and inverse homomorphism, we can conclude that $\operatorname{Duplicate}(L)$ is regular iff $L$ is regular.
