Is an inverse homomorphism always a homomorphism?
If all letters map to the same letter $\delta$, then the inverse $h^{-1}(w) = \{w\mid h(x) = w\}$ is only defined (nonempty) when $w\in\delta^*$ and would consist of all strings in $\Sigma$ that have the same length as $w$. This is not an homomorphism, as these have only one image for each input. (Except for the very special case that $|\Sigma| = |\Delta| = 1$.)

The inverse is a substitution, mapping the letter $\delta$ to the set $\Sigma$  and all other letters in $\Delta$ to the empty set.

Obviously homomorphism $h'$ maps all strings to the empty string. Thus the inverse maps the empty string to $\Sigma^*$ and all other strings to the empty set. Silly mapping, but it is a finite state transduction (finite state automaton mapping with output).  

Let me explicitly answer the question from the title. No. An homomorphism is one-to-one [meaning single valued], an inverse homomorphism in many cases is one-to-many [many-valued]. (If the inverse morphism is one-to-at-most-one [injective] again it usually is not a morphism, but the morphism is called a coding, because it can be "decoded").