Question about a quotient map of a ring

Question

Let $B$ a commutative ring and $I$ an ideal, say $B$ and $B/I$ are both reduced.
Suppose I have a morphism of rings $\phi: B \to B/I$ such that $\phi^{-1}((0)) = I$. Does it then follow that $\phi$ must be the quotient map $b \to b + I$?

Answer 1

No, this does not follow. Consider, for example, $B = \mathbb Z$ and $I = 3 \mathbb Z$. Let $\phi(n) = -n + 3 \mathbb Z$. Then $\phi$ is not the quotient map, but its kernel is $3 \mathbb Z$.

More generally, applying any automorphism of $B$ that preserves $I$ and composing with the quotient map yields another map whose kernel is also $I$.