A minimal automaton for a language quotient $w^{-1}L$ if $L$ is regular
Now that the question is clarified, I transform my comment into an answer and improve the construction. Our goal is to construct a minimal automaton for $w^{-1}L$ given a minimum automaton $A$ for $L$ and $w$.

It is known that $A$ is isomorphic to the automaton where the state correspond to quotients of $L$. This is the minimal automaton that you describe in your question. 

Observe that the quotients of $w^{-1}L$ are also quotients of $L$ since $u^{-1}(w^{-1}L) = (wu)^{-1}L$. Thus, intuitively, you have in $A$ all the states that you need for constructing a minimal automaton for $w^{-1}L$.

First, we claim that the automaton $A$ where we change the initial state to be $w^{-1}L$ recognize $w^{-1}L$. Indeed, by reading $v$ from $w^{-1}L$, you end up in the state $v^{-1}w^{-1}L$. We have then $v \in w^{-1}L$ if and only if $\epsilon \in v^{-1}w^{-1}L$ iff $v^{-1}w^{-1}L$ is final.

Now we have two things to do: first, find the state in $A$ that corresponds to $w^{-1}L$: it is easy, it is the state reached by reading $w$ from $q_0$. Second, we have to minimize this automaton. We claim that if we only keep states that are reachable from state $w^{-1}L$ then we will have a minimal automaton for $w^{-1}L$. Indeed, this automaton still recognize $w^{-1}L$ as we only removed dead state. Moreover each remaining state represents a non-empty quotient of $w^{-1}L$ and two states represent two different quotients. Thus, it has the required minimal size.

To summarize the algorithm:


read $w$ in $A$ and reach state $q$
remove states that can't be reached from $q$
set $q_0 := q$.