State complexity of homomorphisms of regular languages Given a DFA $A = (Q, \Sigma, \delta, q_0, F)$ with $n$ states and a homomorphism $h: \Sigma \to \Gamma^*$. It is easy to see that the family of regular languages is closed under homomorphisms using regular expressions or by construction of a grammar. 

My problem now is to analyze how many states are needed for a DFA $A' = (Q',\Gamma, \delta',q_0',F')$ to accept $h(L)$. First of all I tried to think of a basic construction method to get the automaton.

Let $a \in \Sigma$:  


If $h(a) = \varepsilon$ then all the transitions of $a$ in $A'$ can be removed. 
If $|h(a)| = 1$ then all the transitions with $a$ can be replaced with $h(a)$.
If $|h(a)| \geq 2$ then the transition of $a$ in $A'$ has to be replaced with $h(a)$ and split into $|h(a)|-1$ states each with a one symbol transition

a. If the prefixes of two unique symbols of $\Sigma$ do not share the same prefix then the resulting automaton can be extended into a DFA without adding any states(?).

b. If prefixes are shared then the resulting automaton would be an NFA, therefore it has to be converted into a DFA (using power set construction) and minimizing it.


Therefore my estimation for the state complexity in case $3a$ would be:
$$\begin{align*}|Q'| &amp;\leq |Q| + |Q|\cdot\left(\sum_{a \in \Sigma} |h(a)| - 1\right)\\ &amp;= |Q|\cdot(1- |\Sigma|+\sum_{a \in \Sigma} |h(a)| )\end{align*}$$

However, I'm not convinced about my construction as mentioned in step 3 and was not able to find an example of a language where the upper boundary is met. Therefore I'd be happy to see any advices.
Here is a simple construction that disproves your conjecture.

Let $a,b$ two relatively prime integers. It is known that the largest integer that cannot be represented as a non-negative integer combination of $a,b$ is $ab-a-b$.

Consider now the language $L = (0+1)^*$ and the homomorphism $h$ given by $h(0) = 0^a$ and $h(1) = 0^b$. Then $L$ is accepted by a one-state DFA, while $h(L)$ requires $ab-a-b+1$ states.