(This is inspired by the answer to my earlier question.)
Does there exist
a field $F$ $\:$ and $\:$ two ideals $I$ and $J$ of $F[x]$ $\:$ and $\:$ a ring isomorphism $\: \phi : F[x]/I \to F[x]/J$
such that when $\: q_i : F[x] \to F[x]/I \:$ and $\: q_j : F[x] \to F[x]/J \:$ are the quotient maps,
there does not exist an endomorphism $\: \psi : F[x] \to F[x] \:$ such that $\;\; \phi \circ q_i \: = \: q_j \circ \psi \;\;$?
The answer is yes, if $\phi$ is an isomorphism of rings:
Take $F$ to be the reals, and let $I$ and $J$ both be the ideal of $F[x]$ generated by $x^2+1$. Let $q=q_i=q_j$ be the projection map. The quotient $E:=q(F)$ is isomorphic to the complex numbers. Let $\phi$ be any automorphism of $E$ taking $2^{1/4}$ to $2^{1/4}i$, where $2^{1/4}$ is a real fourth root of 2 and $i$ is $q(x)$.
If $\psi$ is any endomorphism of $F[x]$ then $$q(\psi(2^{1/4}))=q(\pm2^{1/4})=\pm2^{1/4},$$ whereas $$\phi(q(2^{1/4}))=\phi(2^{1/4})=2^{1/4}i.$$ Thus $q\circ\psi\ne\phi\circ q$.
The first equation follows from the fact that $\psi$ preserves $\mathbb{Q}$-conjugates, and also $\psi$ must map $F$ into $F$ because $\psi$ preserves units.
