Identifying the basis a vector is written in Given some vector in a general vector space and the coefficients corresponding to this vector written as a linear combination of some orthonormal basis vectors, is it possible to determine the basis vectors used (to some reasonably unique solution*)? If this is not possible in general, is it possible in $\mathbb{R}^n$ or some other specific collection of vector spaces?
*Reasonably unique in that in some cases, where some of the coefficients are identical, the order may not be determinable, but the set of vectors required for the solution will not change.

As an example, say that I have some vector in $\mathbb{R}^2$, $[2,1]$. Then suppose that I rewrite this vector in terms of the orthogonal basis (I chose the most common basis for simplicity): $[1,0]$ and $[0,1]$, with corresponding coefficients $2$ and $1$. I then provide you with only the original vector $[2,1]$ and the coefficients $2$ and $1$. You know nothing about which basis vectors I have chosen, the only additional information you have is that I used an orthonormal basis. Can you determine which basis vectors I used?
 A: Take your vector $v$, assume it is nr $v$, assume it is non-zero. Then define $x_1:=\frac{v}{||v||}$. Now you may write :
$$\mathbb{R}^n=Vect(x_1)\oplus^{\perp}x_1^{\perp} $$
The vectorial space $x_1^{\perp}$ is of dimension $n-1$. In it the orthonormal bases are in bijection with $O(n-1,\mathbb{R})$. 
For any orthonormal base $(x_2,...,x_n)$ of $x_1^{\perp}$ you see that $(x_1,...,x_n)$ is an orthonormal base of $\mathbb{R}^n$, furthermore in such a base the coordinates of $v$ are always the same : 
$$v_{(x_1,...,x_n)}=(||v||,0,...,0) $$
We have constructed $O(n-1,\mathbb{R})$ different bases for which the coordinates of $v$ is always $(||v||,0,...,0)$... .
A: There are infinitely many such bases. Since orthonormal bases can be identified with the sets of columns of unitaries, your question can be rephrased as follows : given vectors  $x $ and $y $, determine all unitaries $W $ such that $Wx=y $. This is always a big set.
Indeed, If $V $ is another such unitary, then  $Vx=Wx $, or $V^*Wx=x $. So the set we are looking for is of the form $$\{WU:\ Ux=x\}. $$ By considering orthonormal bases where the first element is $x/\|x\|$, our $U $ are of the form  $$U=\begin {bmatrix}1&0\\0& U'\end {bmatrix}, $$ where $U'$ is any $(n-1)\times (n-1) $ unitary.
A: Let's try some geometric reasoning in $\mathbb{R}^3$.
Suppose we have a given vector $\mathbf{r}$ and three real numbers $a$, $b$, $c$, and we know that $\mathbf{r} = a\mathbf{u} + b\mathbf{v} + c\mathbf{w}$, where $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are orthogonal unit vectors. Can we find $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$?
Let $\mathbf{z} = \begin{bmatrix}a&b&c\end{bmatrix}^T$, and form a rotation matrix $\mathbf{M}$ by using $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ as columns. Then we have
$$
\mathbf{r} = 
\begin {bmatrix}
\uparrow & \uparrow & \uparrow\\
\mathbf{u} & \mathbf{v} & \mathbf{w}\\
\downarrow & \downarrow & \downarrow 
\end{bmatrix}
\begin {bmatrix}a\\b\\c\end{bmatrix} = \mathbf{M}\mathbf{z}
$$
So, now the question is this: given $\mathbf{r}$ and $\mathbf{z}$, is there a unique rotation matrix $\mathbf{M}$ that rotates $\mathbf{z}$ to $\mathbf{r}$. Geometrically, the answer is clearly "no". We can take any cone containing $\mathbf{z}$ and $\mathbf{r}$, and some rotation around the axis of this cone will carry $\mathbf{z}$ to $\mathbf{r}$.
You can go through the same reasoning in $\mathbb{R}^2$, and here the answer is that the rotation (i.e. the orthonormal basis) is essentially unique. Given the two vectors $\mathbf{r}$ and $\mathbf{z}$ in $\mathbb{R}^2$, there is essentially only one rotation matrix $\mathbf{M}$ that rotates $\mathbf{z}$ to $\mathbf{r}$. Or, two, at the most.
