For a regular language $L$, is $\{xy^Rz:xyz\in L\}$ regular?
Assume we have automaton $\mathcal A$ for regular language $L(\mathcal A) = L$.
It is possible to construct a new finite automaton for the new language $L'=\{xy^Rz\mid xyz\in L\}$. You need nondeterminism and extra components in the states to remember the choices made.

It is also possible to use closure properties. Let $\mathcal A_{p.q}$ be the automaton $\mathcal A$, except that initial state is $p$ and final state is $q$. Similarly $\mathcal A_{p.F}$ for final states $F$.

Now $L'= \bigcup_{p,q} L(\mathcal A_{q_0,p})\cdot L(\mathcal A_{p,q})^R\cdot L(\mathcal A_{q,F})$.