To prove uniqueness of the rotation tensor associated with rotation of a rigid body Suppose there are $N$ particles embedded in a rigid body which undergoes some random rotation such that: 
$$ \overline{\overline {R}}_{ij} \otimes \vec{a}_{ij} =  \vec{b}_{ij}$$
where, 


*

*$i$ and $j$ are just two random particles out of the $N$ particles

*$\vec{a}_{ij}$ is a vector connecting the $i^{th}$ and the $j^{th}$ particle

*$\vec{b}_{ij}$ is the rotated version of $\vec{a}_{ij}$

*$\overline{\overline {R}}_{ij}$  is the rotation tensor responsble for the rotaion of $\vec{a}_{ij}$ to $ \vec{b}_{ij}$


I intend to prove that $\overline{\overline {R}}_{ij}$ remains the same for all the combinations of $i$ and $j$ belonging to the set $[1,N]$ for $(i \neq j)$ using the most fundamental definition of a rigid body that it is a collection of particles which remains equidistant from each other.
An Update on my effort:
Using the definition of a rigid body - the distance between any two particles remains constant, we can also say that - the dot product of any two vectors joining the particles embedded in the rigid body remains the same before and after the rotation. 
So, taking into account the $i^{th}, j^{th}, k^{th}$ and the $l^{th} $ particle, we can write -
the dot product of the vectors before the rotation = the dot product of the two vectors after rotation
$$ \vec{a}_{ji} \cdot \vec{a}_{lk}=(\overline{\overline {R}}_{ji} \otimes \vec{a}_{ji}) \cdot (\overline{\overline {R}}_{lk} \otimes \vec{a}_{lk})$$ 
Now, using the canonical isomorphism  property of  the tensor product:
$$ \vec{x} \cdot (\overline{\overline Z} \otimes  \vec{y}) = \vec{y} \cdot (\overline{\overline {Z}}^T \otimes  \vec{x})$$
We have,
$$ \vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{lk} \cdot [\overline{\overline {R}}^T_{lk} \otimes  (\overline{\overline {R}}_{ji} \otimes \vec{a}_{ji}) ]$$
Using the associative property of tensor product:
$$\vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{lk} \cdot (\overline{\overline {R}}^T_{lk} \otimes  \overline{\overline {R}}_{ji}) \otimes \vec{a}_{ji} $$
Then again using the canonical isomorphism  property, we have:
$$ \vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{ji} \cdot (\overline{\overline {R}}_{lk} \otimes  \overline{\overline {R}}^T_{ji}) \otimes \vec{a}_{lk} $$
Which implies,
$$\vec{a}_{ji} \cdot \{ [\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes  \overline{\overline {R}}^T_{ji})] \otimes \vec{a}_{lk} \}=0$$
Now, given the fact that the chosen vectors, $\vec{a}_{ji}$ and $\vec{a}_{lk}$, are non zero, the left hand side could only become zero under the following conditions:


*

*Either the last tensor product is zero

*Or the first dot product is zero

*Or the tensor, $[\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes  \overline{\overline {R}}^T_{ji})]$, itself is zero

*Or any combination of the above three  mentioned conditions are true


Now if somehow I prove that the first, the second and the fourth points are wrong, I'll be left with only one possibility that, $[\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes  \overline{\overline {R}}^T_{ji})]=0$, proving  which also happens to be my primary objective right now. And this is where I needed help. If what I asked for is proved, the rest of the task is pretty much tractable.
Thank you.
 A: Your Problem
If you're trying to prove that all $R$ must be equal in order for all the distances to remain the same after rotation, you will fail, as it's not true. Rotation $R_{ij}$ could be replaced by any other rotation that only differs by a rotation about $a_{ij}$ for example say you have points $i=\langle 0,0,0 \rangle$ $j=\langle 1,0,0 \rangle$ and all $R$ were
$\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 1 \end{bmatrix}$ then $R_{ij}$ could be any matrix such that $R_{ij} \otimes \langle 1,0,0 \rangle = \langle -1,0,0 \rangle$ and any matrix of the form $\begin{bmatrix}
0 & a & b \\
-1 & c & d \\
0 & e & f \end{bmatrix}$ would suffice. This is a trivial example to demonstrate the point but the same concept  it true in general.
What I think you're actually interested in
Prove that for a set of points with coordinates $p_i$ which are arbitrarily displaced to coordinates $p'_i$ that if the distances between the each pair of points remains constant across the displacement, that there exists a unique $R$ such that $R\otimes p_i=p'_i$
And to get you started with the proof you can use $|p_i-p_j|=|p'_i-p'_j|$ as the distance constraint, which can be rewritten as $(p_i-p_j)\cdot(p_i-p_j)=(p'_i-p'_j)\cdot(p'_i-p'_j)$
One thing that you will undoubtedly run into, is that the number of points matters. You must have at least $n\frac{n-1}2$ points, where n is the number of dimensions you have, for the unique constraint to hold true.
