what are the conditions for two NFAs to be equivalent?

Question

It's a pretty basic question, but I couldn't find an explicit answer in the book which I'm using, also when searching for it - I didn't find a reliable reference.

Correct me if I'm wrong but for two DFA's, $A , A'$ -given that $L(A) = L(A')$ doesn't mean that the DFA's are equivalent, so what are the (sufficient and necessary) conditions under which $A$ and $A'$ are equivalent? The closest I found was a claim states that in case $\bigcup_{q\in Q} L(q) \neq \bigcup_{q'\in Q'}L(q')$ (when $L(q) = \{w\in\Sigma^* : \hat{\delta}(q_0,w) =q \})$ then $A, A'$ are not equivalent.

Answer 1

As commented, they are equal when they accept the same language.

In practice:

Every NFA has one (1) minimized DFA, which is unique. If two minimized DFA are equal, the corresponding NFA's must also be equal.