Does an isomorphism between quotient rings deduce an isomorphism between rings? Exactly I mean if this proposition is true or in what case it stays true:

$I_1$ and $I_2$ are ideals of $R$, then $R/I_1\cong R/I_2$ if and only if $\exists\varphi\in Aut(R)$, such that $\varphi(I_1)=I_2$

The if part is naturally deduced by applying the First Isomorphism Theorem to $\pi\circ\varphi$, where $\pi$ is the canonical map from $R$ to $R/I_2$, but the only if part puzzles me.
If the only if part is true, how to construct an automorphism?
If it is false, what is the counterexample? Can I restrict $R$ to some kinds of rings so that the proposition stays true?(e.g. PIR, I tried to solve the problem in this case, because an automorphism maps $a$ to $b$ naturally maps $(a)$ to $(b)$, but I failed to get further)
 A: This is false, even for PIDs.  Let $R = \mathbf Z[i]$, $I_1 = (4+7i)$ and $I_2 = (1+8i)$.  Note $4^2 + 7^2 = 1^2 + 8^2 = 65$, $4+7i = (2+i)(3+2i)$, and $1+8i = (2+i)(2+3i)$.  The factors $2+i$, $3+2i$, and $2+3i = (i)(3-2i)$ are all prime in $R$ and by the Chinese remainder theorem
$$
R/I_1 \cong R/(2+i) \times R/(3+2i) \cong \mathbf Z/(5) \times \mathbf Z/(13) \cong \mathbf Z/(65)
$$ and
$$
R/I_2 \cong R/(2+i) \times R/(2+3i) \cong \mathbf Z/(5) \times \mathbf Z/(13) \cong \mathbf Z/(65),
$$
so $R/I_1$ and $R/I_2$ are isomorphic rings. The only ring automorphisms of $R$ are the identity and complex conjugation.  These automorphisms map the ideal $I_1 = (4+7i)$ to $(4+7i)$ and $(4-7i)$, neither of which is $(1+8i)$.
There is nothing that special about $\mathbf Z[i]$ here: you can build similar counterexamples with lots of quadratic rings.  The point is that you can have a lot of numbers in a quadratic ring with the same norm even though the numbers are not unit multiples of each other. In the case of $\mathbf Z[i]$ this amounts to saying there are positive integers that can be a sum of two squares $a^2 + b^2$ in lots of essentially different ways (not just from permuting the roles of $a$ and $b$ or making sign changes on $a$ or $b$).
The lesson here is that passing to quotient rings involves a lot of collapsing, so you should not really expect that the "only" way quotient rings of $R$ can be isomorphic in general should be from reducing some automorphism of $R$.
