You have that if $A \leftrightarrow B$, and $A \leftrightarrow C$, then $B \leftrightarrow C$. So, if $B$ is necessary and sufficient for $A$, and if $C$ is necessary and sufficient for $A$ as well, then $B$ and $C$ will be necessary and sufficient for each other as well.
So, the fact that $B$ and $C$ are 'different' is only superficial. Yes, you can always rephrase a set of necessary and sufficient conditions in such a way that it can be said to be a different set of necessary and sufficient conditions, but the 'difference' would be merely syntactical. So yes, $A \leftrightarrow B$ does not imply $A \leftrightarrow C$, but that is because by using different letters $B$ and $C$ for the two sets of necessary and sufficient conditions you are simply hiding their actual equivalence behind syntactical symbols. And indeed, if you throw their actual equivalence back in, i.e. If you add $B \leftrightarrow C$, then from $A \leftrightarrow B$ you now can infer that $A \leftrightarrow C$
Consider your dialogue. First, as Mauro points out, if paying an advance of 3 months rent is a necessary condition, then if that is not part of the initial signing of the lease and agreeing to pay $600$ dolars per month, then person $A$ was simply lying when saying that the intial agreement was a necessary and sufficient condition ... the initial agreement would merely have been a necessary condition, to which one could indeed add any number of others. So, for your story to work, it must have been the case that paying 3 months of rent in advance was simply part and parcel of that initial agreement: presumably it is spelled out in the agreement that the renter has to pay 3 months in advance, and that it is in fact that very part of the initial agreement that, to person A, is necessary and sufficient for B to get the keys. And so, we are not talking about completely sets of necessary and sufficient conditions after all.
No, because the sufficient property could be false: if $n=p=\top,s=\bot$, then $p⇒ n$ and $s⇒p$ are both true, but $n\wedge s=\bot$ while $p$ remains true, so the equivalence certainly cannot hold.
One could see that on an intuitive level by noting that since $s\implies p\implies n$, the statement $s\wedge n$ only depends on the statement of the stronger of both, namely the sufficient condition $s$. But that would reduce the right side to $s\Leftrightarrow p$, which is absurd in general.
Best Answer
We say that B is a necessary condition for A to mean: $A \rightarrow B$.
And we say also that A is a sufficient condition for B to mean: $A \rightarrow B$.
Let $P := (x \to y) \to z$ and $Q := x \to (y \to z)$ and we have to check if $Q$ is a necessary/sufficient condition for $P$.
Thus Q is a sufficient condition for P iff $Q \to P$ is true and Q is a necessary condition for P iff $P \to Q$ is true.