What is the significance of being equivalent up to local isometry? Background :
I am reading the paper device independent outlook on quantum mechanics. The author mentions the concept of two pure states being equivalent up local isometry. From what I understood two states $|s\rangle$ and $|t\rangle$ are equivalent up to local isometry if by having the option of appending extra degree of freedom to $|s\rangle$ and performing local unitary operations I can extract out ( the meaning of extract out becomes clear in the example below ) state $|t\rangle$. For example consider the state ( here $|k\rangle$ are orthonormal vectors ) $$|\psi\rangle_{AB}=\sum_k c_k\frac{1}{\sqrt{2}}(|2k,2k\rangle+|2k+1,2k+1\rangle)$$
Let the unitary $U=U_A \otimes U_B$ where $U_A$ is such that it acts in the following way  ( $|00\rangle$ is the ancilla  ) 
$$U_A|2k,0\rangle = |2k,0\rangle$$
$$U_A|2k+1,0\rangle = |2k,1\rangle$$
$U_B$ acts in the same way. Now if I append the extra degree of freedom $|00\rangle$ to $|\psi\rangle_{AB}$ and apply $U$
$$U|\psi\rangle_{AB}|00\rangle_{AB} = |\phi\rangle \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$
Here $|\phi\rangle $ is not of interest to me. But by appending the extra degree of freedom and applying the local unitary operations I was able to extract state $|\phi^{+}\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. Thus $|\psi\rangle_{AB}$ is equivalent to $|\phi^{+}\rangle$ upto local isometry.

Doubts:


*

*Is my understanding of local isometry correct ?

*If yes, then is being equivalent up to local isometry symmetric ( I doubt it,but is answer is no the relations becomes wierd to me ). I mean if $|s\rangle$ is equivalent to $|t\rangle$ , is the reverse also true ?

*The example above I gave was a specific one from the paper itself. But in general how 
do I prove that if $|s\rangle$ is equivalent to $|t\rangle$ then for every measurement settings on state $|t\rangle$ there will be some measurement settings for state $|s\rangle$ which gives the same statistics ?

 A: Let's start with questions 1, for which the answer is "no".
An isometric quantum channel $T_{A\to A^{\prime}}$ is a channel $T_{A\to A^{\prime}}:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}^{\prime}_A)$ such that $T_A(\rho)=U\rho U^{\dagger}$ with $U^{\dagger}U=1$ and $UU^{\dagger}=P$ with some projection onto a subspace. The conditions on $U$ mean that the channel is an isometry.
A local isometric channel is a channel that acts locally (as in: in the form of products) on a multipartite system, i.e. for a bipartite setting, a local isometric channel would be of the form $T_{A\to A^{\prime}}\otimes T_{B\to B^{\prime}}:\mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)\to \mathcal{B}(\mathcal{H}_A^{\prime}\otimes \mathcal{H}_B^{\prime})$.
Similarly, I would define that a given state $\rho\in \mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)$ is equivalent to another state $\sigma\in \mathcal{B}(\mathcal{H}_A^{\prime}\otimes \mathcal{H}_B^{\prime})$ via local isometries if there exists a local isometric channel $T_{A\to A^{\prime}}\otimes T_{B\to B^{\prime}}$ that sends $\rho$ to $\sigma$ and there is a local isometric channel $T^{\prime}_{A^{\prime}\to A}\otimes T^{\prime}_{B^{\prime}\to B}$ sending $\sigma$ to $\rho$ (otherwise it wouldn't be an equivalence - which also answers your second question).
Now, one class of local isometries is of course given by local unitaries, but I guess the claim is that there are also states that are equivalent up to local isometries but that are not equivalent up to local unitaries (otherwise the idea would be rather senseless). I guess, this is where measurements on ancillary systems come into play, but I don't really know.
What the paper you read seems to imply is that the concept of local isometries is the right concept to describe states that look the same if only the (local) measurement statistics described there are available. It might be that this entails what you understood as the concept of being equivalent up to local isometries, but I didn't do the math.
