Action of GL(2,O_k) on 1d subspaces of (O_k)^2

Question

Let $\mathcal{O}_k$ be the ring of integers in an algebraic number field $k$. Let $M$ be a rank $1$ projective module over $\mathcal{O}_k$ (in other words, $M$ is a projective module such that $k \otimes M \cong k$). Let $X_M$ be the set of all direct summands of $(\mathcal{O}_k)^2$ which are isomorphic to $M$. Using the classification of finitely generated projective modules over Dedekind domains, we see that $X_M$ is nonempty. The group $\text{GL}_2(\mathcal{O}_k)$ acts on $X_M$.

Question : Does $\text{GL}_2(\mathcal{O}_k)$ act transitively on $X_M$?

I should remark that this is true if $M \cong \mathcal{O}_k$. Indeed, let $L \in X_{\mathcal{O}_k}$. Then there exists some $\vec{v} \in (\mathcal{O}_k)^2$ such that $M = \mathcal{O}_k \cdot \vec{v}$, and writing $\vec{v} = (v_1,v_2)$ we have that $\mathcal{O}_k \cdot v_1 + \mathcal{P}_k \cdot v_2 = \mathcal{O}_k$. We can thus find $w_1,w_2 \in \mathcal{O}_k$ such that $w_1 v_1 + w_2 v_2 = 1$, and hence the matrix $\left(\begin{smallmatrix} v_1 & -w_2 \\ v_2 & w_1 \end{smallmatrix}\right)$ lies in $\text{SL}_2(\mathcal{O}_k)$ and takes $(1,0)$ to $\vec{v}$.

Answer 1

I think this is clear; the complementary summand to M will be its inverse M' in the Picard group, so if you have have a second such summand N, you just pick isomorphisms $M\cong N$ and $M'\cong N'$, and this will induce an automorphism of $\mathcal{O}_k^2$.