Is an overring of an order reflexive as a module over the order?

Question

Let $R$ be a one-dimensional Noetherian domain with fraction field $K$, let $\tilde{R}$ be the integral closure of $R$ in $K$, and assume that $\tilde{R}$ is finitely generated as an $R$-module. (In this situation one sometimes says that $R$ is an "order" in the Dedekind domain $\tilde{R}$.) For a fractional $R$-ideal $I$ we have

$(R:I) := \{x \in K \mid x I \subset R\} = \operatorname{Hom}_R(I,R)$.

Question 1: Does it always hold that $(R:(R:\tilde{R})) = \tilde{R}$?

Question 2: Same question but with an intermediate ring $R \subset R' \subset \tilde{R}$. Does it always hold that $(R:(R:R')) = R'$?

Comments:

1) For any fractional $R$-ideal $I$ we have

$(R:I) = \operatorname{Hom}_R(I,R)= I^{\vee}$.

Therefore my questions are equivalent to asking whether $\tilde{R}$ and $R'$ must be reflexive as $R$-modules. A one-dimensional Noetherian domain $R$ is Gorenstein iff every fractional ideal is reflexive, so the answer to 2) is yes if $R$ is Gorenstein.

2) In a draft of a paper of mine, I wrote down $(R:(R:R')) = R'$ without any justification. Now I'm concerned! The main results of the paper apply to a class of orders that are in fact Gorenstein, so it's not so terrible, but I would like to extend these results to a larger class of orders if possible (and also build up background results in a graceful way).

Answer 1

The answer to Question 1 is yes by (the proof of) Proposition 2.14 in this paper.

The answer to Question 2 is not. Let $R=k[t^3,t^4,t^5]$ so $\tilde R=k[t]$. I claim that for this ring any immediate ring $R\subsetneq R'\subsetneq \tilde R$ is not reflexive.

Let $m=(t^3,t^4,t^5)$ be that maximal ideal of $R$. As the semigroup generated by $(3,4,5)$ contains all numbers from $3$, $mR'\subseteq m\tilde R\subset R$. It follows that $Hom_R(R',R)$ contains $m$, and since it has to be an ideal of $R$ and can not be $R$, it is $m$.

So now we just need to show that $Hom_R(m,R)\neq R'$. But this is clearly equal to $\tilde R$, as multiplication by $t^i, i\geq 0$ maps $m$ to $R$ and no negative power works.

In general, I think for a non-Gorenstein domain of dimension one, torsion free modules are rarely reflexive.