Let $K$ be a totally real cubic number field of dicriminant $d$, in this article the selmer group of $K$ is defined as
$$\text{Sel}(K):=\{\alpha \in K^{\times}: (\alpha)=\mathfrak{a}^2 \}/K^{\times\,2}.$$ We also have the subgroup of totally positive elements, given by
$$\text{Sel}^{+}(K):=\{\alpha \in K^{\times}: (\alpha)=\mathfrak{a}^2 ,\alpha \gg 0\}/K^{\times\,2}.$$
Now if $\mathcal{C}(d)$ is the set of rank three positive definite integral lattices of discrimant $d$ up to isometry ( which is the same as the set of positive definite ternary quadratic forms up to the action of $GL_3(\mathbb{Z})$), then we have a natural map
$$\psi:\text{Sel}^{+}(K) \to \mathcal{C}(d)$$
defined by $\psi \left(\alpha K^{\times\,2} \right)=(\mathfrak{a},q_{\alpha})$ where $(\alpha)=\mathfrak{a}^2$ and $q_{\alpha}(x)=\text{tr}_{K/\mathbb{Q}}(\frac{x^2}{\alpha})$ for all $x \in \mathfrak{a}$. This map is well defined, because if $\beta K^{\times\,2} =\alpha K^{\times\,2}$ and $\beta=\alpha \cdot \gamma^2$, then $\phi:(\mathfrak{a},q_{\alpha}) \to (\mathfrak{\gamma a},q_{\beta})$ given by $x \mapsto \gamma \cdot x$ is an isometry between $\psi \left(\alpha K^{\times\,2} \right)=(\mathfrak{a},q_{\alpha}) $ and $\psi \left(\beta K^{\times\,2} \right)=(\mathfrak{\gamma a},q_{\beta})$.
Moreover, in this previous comment it's (basically) proved that if $\psi \left(\alpha K^{\times\,2} \right)=\psi \left(1 \cdot K^{\times\,2} \right)$, then $\alpha \in K^{\times\,2} $, which make tempting to conjecture that $\psi$ is in fact injective, however this fails miserably for cyclic cubic fields, as the example in the same comment shows. All of this led to following question:
If $K$ is a non-cylic cubic field, then is $\psi$ injective?
One idea to prove this (based in the proof of the comment and in the existing examples) would be to suposse $\phi:(\mathfrak{a},q_{\alpha}) \to (\mathfrak{b},q_{\beta})$ is an isometry, i.e., is an isomorphism of abelian groups such that $\text{tr}_{K/\mathbb{Q}}(\frac{\phi(x)^2}{\beta})=\text{tr}_{K/\mathbb{Q}}(\frac{x^2}{\alpha} )$ for all $x \in \mathfrak{a}$, if we could prove that this implies that for some $x \in \mathfrak{a}$, the elements $\frac{x^2}{\alpha}$ and $\frac{\phi(x)^2}{\beta}$, have the same minimal polynomial then they would be equal (since $K$ is non-cyclic) and thus $\beta K^{\times\,2} =\alpha K^{\times\,2}$ , however it's been impossible to my to find a way to do that.
I hope the post wasn't too long, any help will be appreciated.