finite automata under morphism Given two (deterministic) finite automata $A, B$ over $\Sigma$, a mapping $h:\Sigma\rightarrow \Sigma'$

Naturally $h$ can be extended to a mapping in $\Sigma^*\rightarrow \Sigma'^*$ which is denoted by $h$ as well. 

Is the set 
$$\{w\in L(A)\mid h^{-1}(h(w))\subseteq L(B)\}$$ regular?
The answer is yes. To prove this, consider the complement language:
$$\{w: w\notin L(A) \vee \exists x: h(x)=h(w)\wedge x\notin L(B)\}$$

Now, we construct an NFA $C$ as follows. Given a word $w$, $C$ has $\overline{A}$ as a component, such that if $w\notin L(A)$ then $C$ accepts. It remains to check the rest.

Let $D$ be a DFA that recognizes $\overline{B}$.
In order to check if there exists a word $x$ as described, $C$ guesses (i.e. has a nondeterministic transition), for every letter $\sigma$ of $w$, a transition in $D$ that can be taken with a letter $\tau$ such that $h(\tau)=h(\sigma)$. The accepting states of this component are those of $D$.

It is not hard to prove that $C$ accepts the language above, and by complementing it you get an automaton for your desired language.