Smallest collection of linear operators satisfying isometry property

Let $$\mathscr{A}=\{(A_{1,1},A_{1,2},A_{1,3}),...,(A_{S,1},A_{S,2},A_{S,3})\}$$ be a collection of linear operators $$A_{n,k}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$$. For $$u$$ and $$v\in (\mathbb{R}^2)^3$$, if $$\|\sum\limits_{k=1}^{3}A_{n,k}u_{k}||^2=||\sum\limits_{k=1}^{3}A_{n,k}v_{k}\|^2 \text{ for } n=1,...,S$$ then $$u$$ and $$v$$ will be said to be $$\mathscr{A}$$-equivalent, denoted by $$u\simeq_{\mathscr{A}}v$$. Call an isometry $$T:\mathbb{R}^2\rightarrow\mathbb{R}^2$$ $$\mathscr{A}$$-admissable if $$T$$ commutes with all operators in the tuples contained in $$\mathscr{A}$$ (that is, $$T A_{n,k}=A_{n,k}T$$, so that $$\|\sum\limits_{k=1}^{3}A_{n,k}Tu_{k}||^2=\|\sum\limits_{k=1}^{3}A_{n,k}u_{k}||^2$$).

Question: What is the smallest sized collection of linear operators, $$\mathscr{A}=\{(A_{1,1},A_{1,2},A_{1,3}),...,(A_{S,1},A_{S,2},A_{S,3})\},\ |\mathscr{A}|=S,$$ such that for any pair of tuples $$u,v\in(\mathbb{R}^2)^3$$ $$\|\sum\limits_{k=1}^{3}A_{n,k}u_{k}||^2=\|\sum\limits_{k=1}^{3}A_{n,k}v_{k}\|^2 \text{ for all }n=1,...,S$$ implies $$u$$ and $$v$$ belong to the same orbit of some $$\mathscr{A}$$-admissable isometry, $$T$$, that is $$Tu_{k}=v_{k}$$ for all $$k=1,2,3$$ for some $$\mathscr{A}$$-admissable isometry $$T$$?

Remark: The size has to be at least $$5$$ but no more than $$6$$. The upper bound follows from the argument below.

Let $$u,v \in (\mathbb{R}^2)^3$$ be given. Consider the collection $$\mathscr{A}$$ consisting of $$3$$-tuples of operators $$E_1=(I,0,0),E_2=(0,I,0),\ \text{ and } E_3=(0,0,I)$$ and the $$3$$-tuples $$F_{1,2}=(I,I,0),F_{1,3}=(I,0,I),\text{ and }F_{2,3}=(0,I,I)$$. Then the condition $$\|\sum\limits_{k=1}^{3}A_{n,k}u_{k}||^2=||\sum\limits_{k=1}^{3}A_{n,k}v_{k}\|^2 \text{ for } n=1,...,6$$ implies existence of isometry $$T$$ such that $$Tu_{k}=v_{k}$$ for all $$k=1,...,3$$ (intuitively, knowing the norms of the action of these operators tells all of the inner products between elements) and $$T$$ commutes with all of the members of $$\mathscr{A}$$ as they are just identities.

The lower bound depends on the fact that the manifold $$\mathbb{R}^{MN}$$ when quotient out by the closed sub-group $$H=O(N,\mathscr{A}) = \{T\in O(N) | T A_{n,k}=A_{n,k}T, A_{n,k}\in\mathscr{A}\}$$ of $$O(N)$$ has dimension as a quotient space equal to that of the principle orbits $$\dim(\mathbb{R}^{MN}/H)\geq \dim(\mathbb{R}^{MN}/O(N))= MN-\dim(O(N))=3(2)-1=5$$. Thus the mapping $$\psi :(\mathbb{R}^2)^3/O(2,\mathscr{A})\rightarrow \mathbb{R}^{S}$$ $$[(u_1,u_2, u_3)] \mapsto (\| \sum\limits_{k=1}^3 A_{1,k}u_k \|^2, ..., \| \sum\limits_{k=1}^3 A_{S,k}u_k \|^2)$$ which is continuous can be injective only if the dimension of $$S$$ is at least $$5$$ (by invariance of domain).

Note: Some may recognize some similarities between this type of problem and its motivating problem of phase retrieval, where the absolute values of measurements $$|\langle x, \varphi_{n} \rangle|$$ of a signal $$x$$ can be observed, and recovery is only up to a plus or minus multiple of a signal $$x\in\mathbb{R}^M$$. There (in the real case) it is known precisely conditions for the minimal number of operators. These conditions are characterized by the measuring operators satisfying what is known as the complement property holding for $$\{\phi_n\}_{n=1}^S$$, and precisely $$S=2M-1$$ measurements suffices.