Is the inverse of a finitely generated $A$-submodule of $\mathrm{Frac}(A)$ still finitely generated?

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Context. I recently realized that the definition of a fractional ideal of an integral domain $A$ I knew was probably the wrong one.

To give some context, let me give two different possible definitions of a fractional ideal.

Let $A$ be an integral domain, with field of fractions $K$.

Definition 1. ("wrong one") A fractional ideal of $A$ is a finitely generated $A$-submodule of $K$.

Definition 2. ("right one") A fractional ideal of $A$ is an $A$-submodule of $K$ of the form $x\mathfrak{a},$ where $x\in K^\times$ and $\mathfrak{a}$ is an ideal of $A$.

Recall that if $I$ is an $A$-submodule of $K$, then $I^{-1}=\{x\in K\mid xI\subset A\}$. We say that $I$ is invertible if $II^{-1}=A$.

Clearly, a fractional ideal in the sense of Definition $1$ is also a fractional ideal in the sense of definition $2$. The opposite direction is not necessarily true if $A$ is not noetherian.

It is also known that an invertible submodule $I$ is automatically finitely generated, so the two definitions coincide for invertible fractional ideals.

Definition 2 seems more tractable since one may prove easily that if $I$ is a fractional ideal in the sense of Definition $2$, then the same holds for $I^{-1}$.

Here comes finally my question:

Question. Let $I$ be an arbitrary finitely generated $A$-submodule of $K$. Is $I^{-1}$ necessarily finitely generated? If not, what about if we assume $A$ noetherian?

(In other words, is the inverse of a fractional ideal in the sense of Definition 1 still a fractional ideal in the same sense?)

I think the answer should be NO but I can't cook up any counterexample for the moment.

Thank you in advance for your insights.

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Let $k$ be a field and consider the ring $A=k[x,xy,xy^2,xy^3,\dots]\subset k[x,y]$ and the $A$-submodule $I\subset k(x,y)$ generated by $y$ and $1$. Then $I^{-1}$ is the ideal in $A$ generated by $x,xy,xy^2,\dots$ which is not finitely generated.