For an ideal $\mathfrak{a}$, find an $x$ such that $\mathfrak{d}(x)=\mathfrak{a}^{-1}$

Question

Let $K$ be a number field, $\mathfrak{a}$ a nonzero ideal of $\mathcal{O}_{K}$, and $\mathfrak{d}$ the different. If $\mathcal{O}_{K}$ is a PID, then one can find an $x$ such that $\mathfrak{d}(x)=\mathfrak{a}^{-1}$.

Is it possible to do this for any $K$ (in which the different may also not be principal)?

Answer 1

No, since the existence of $x$ implies that $\mathfrak{a}\mathfrak{d}$ is a principal ideal.

If you want a counterexample, take any number field $K$ where the different ideal is not principal and $\mathfrak{a}=\mathcal{O}_K$.