DFA's with more than one initial state
As Lukas points out, having multiple initial states is something of a red herring. You've defined your automaton to accept a subset of the strings that match the regular expression $Q\Sigma^*$ and that regular expression matches the language accepted by some ordinary DFA using the usual techniques. (Lukas gives an explicit construction in terms of the original automaton.)

For equivalence, there is a standard polynomial-time technique used for state minimization for DFAs. Luca Trevisan has a good set of notes (PDF) on this, which I'll briefly summarize.  Let the automaton have state set&nbsp;$Q$, alphabet&nbsp;$\Sigma$ and transition function&nbsp;$\delta$.

For a state $q$ and string&nbsp;$w$, write $\delta^*(q,w)$ for the state that you reach by starting at&nbsp;$q$ and reading&nbsp;$w$.  For states $p$ and&nbsp;$q$, write $p\equiv q$ to mean that $\delta^*(p,w)$ and $\delta^*(q,w)$ are either both accepting or both rejecting. This gives the same relation&nbsp;$\equiv$ that you give in the question.  We'll show how to compute&nbsp;$\equiv$.

To do this, for any $n\geq 0$, define $p\equiv_n q$ in the same way as $p\equiv q$, except restricting $w$ to be of length at most&nbsp;$n$.  We'll compute the relations&nbsp;$\equiv_n$ inductively, and show that knowing a finite number of them is the same as knowing&nbsp;$\equiv$.

First, note that $p\equiv_0 q$ if, and only if, $p$ and&nbsp;$q$ are both accepting or both rejecting so we can compute&nbsp;$\equiv_0$ easily.  Now, suppose we've already computed&nbsp;$\equiv_n$.  We claim that $p\equiv_{n+1}q$ if, and only if, $p\equiv_n q$ and $\delta(p,u)\equiv_n\delta(q,u)$ for all $u\in\Sigma$. Obviously, if $p$ and&nbsp;$q$ "agree" on all strings of length at most $n+1$, they must also agree on all strings of length at most&nbsp;$n$. For the second part, consider a string $uw$ where $|u|=1$ and $|w|=n$. We're saying that, for $p$ and&nbsp;$q$ to agree on $uw$, it must be that, after reading&nbsp;$u$, they move to states that agree on&nbsp;$w$.

This allows us to construct the relations $\equiv_n$ for all&nbsp;$n$.  Finally, we claim that $\equiv_{|Q|}$ is actually the same relation as&nbsp;$\equiv$.  To see this, observe that every $\equiv_n$ is an equivalence relation and, furthermore, the procedure from moving from $\equiv_n$ to&nbsp;$\equiv_{n+1}$ can only split equivalence classes into smaller ones: it can never merge them. Also, if $\equiv_n$ and&nbsp;$\equiv_{n+1}$ are the same relation, then $\equiv_{n+i}$&nbsp;is the same relation for all $i\geq 0$.  $\equiv_0$&nbsp;has at most two equivalence classes.  Every step of the process either splits at least one equivalence class or is the end of the game: it doesn't split any class and nor does any future step. We can't split classes more than $|Q|$&nbsp;times because, if we did, we'd have more equivalence classes than states, which is impossible. So it must be that $\equiv_{|Q|}$ is the same thing as&nbsp;$\equiv$ and we're done.