Describing an arbitrary transformation of two vectors which preserves their lengths & angles I'm trying to come up with a compact metric which describes the similarity between two pairs of vectors in 3D (or higher) space, assuming that the angle between the two vectors in each pair is the same. Something that seems logical is angle. However, I have an intuition that one angle will not suffice for describing this relationship. 
My reasoning is this: I (think I) know that you can describe any length-preserving transformation of a single vector in 3D in terms of an axis of rotation and an angle. However, if there is a second vector, this vector should be free to be oriented in any direction around the first (after its rotation), still  maintaining the same angle.
A concrete example: vector pair 1 is $\{(1,0,0),(0,1,0)\}$. I can rotate both $90^\circ$ around $(1,0,0)$ to get $\{(1,0,0),(0,0,1)\}$, then around $(0,1,0)$ to get $\{(0,0,1),(-1,0,0)\}$. But is there any way to get from $\{(1,0,0),(0,1,0)\}$ to $\{(0,0,1),(-1,0,0)\}$ in a single rotation? If not, what is a rigorous way to show this?
I have a colleague who suspects that this can be done if we use additional dimensions.
 A: You’re almost there. A composition of rotations is a rotation, so there’s no need to go outside of $\mathbb R^3$. Add the cross product of each pair of vectors to each set and assemble them into $3\times3$ matrices, so that you have $V=\begin{bmatrix}\mathbf v_1 &\mathbf v_2&\mathbf v_1\times\mathbf v_2\end{bmatrix}$ and $W=\begin{bmatrix}\mathbf w_1&\mathbf w_2&\mathbf w_1\times\mathbf w_2\end{bmatrix}$. Assuming that the original vectors were not colinear, you’ve now got two bases for $\mathbb R^3$. The transformation that maps one onto the other is $R=WV^{-1}$. If the original lengths and angles were matched as you say, this will be a rotation.  
Using your example, we have $$V=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I, W=\begin{bmatrix}0&-1&0\\0&0&-1\\1&0&0\end{bmatrix}$$ and so $$R=WV^{-1}=W.$$ This is a rotation about $(1,-1,1)^T$ through an angle of $$\cos\theta = \frac12(\operatorname{tr}R-1)=-\frac12,$$ i.e., $\theta=\frac23\pi$.
A: The linear transformations that preserve lengths and angles between vectors are exactly the orthogonal transformations. If you also restrict yourself to orientation preserving transformations (which is no problem here, since you only deal with two vectors), you exactly get the group of rotations $SO(n)$. Now in three dimensions, you have Euler's rotation theorem, which actually tells you that any combination of rotations can again be described by a single rotation.
The other answer already gave you the calculation while I was typing, but I hope I could give you some additional hints on what to search for.
By the way, interestingly your colleague was doubly wrong. Not only is this possible in 3D, Euler's rotation theorem does not hold for higher dimensions. Let's say I have a 4D vector (w,x,y,z), which I first rotate by $90°$ in the w-x-plane (there are no rotational axes in 4D, we can only rotate in a certain plane instead) and then in the y-z-plane. Then there is no way to express this as a single rotation.
