Can different quaternions represent the same orientation? For a current project I'm working on I  have to use quaternions to represent the orientation of an object. The piece of code I'm working on now integrates rotation rates to the quaternion representing the orientation. To test it I made a simulink model to simulate a rotating object and compare my result with the one from the simulation. 
I noticed that if I compare the quaternion calculated by Simulink's "rotation angles to quaternion" block, it's not the same as the one I calculated, but if I convert both back to rotation angles they are the same.
This leads me to the conclusion that several quaternions can represent the same orientation. But since I'm new to quaternions I thought I might just ask to be sure. 
 A: If a quaternion is multiplied by any real number, it still represents the same rotation in $\mathbb{R}^3$.  That's the only way two quaternions can represent the same rotation.
In more detail: $\mathbb{R}^3$ can be identified with the quaternions $E = \{ xi + yj + zk : x,y,z \in \mathbb{R}\}$ having zero real part, via the obvious map $(x,y,z) \mapsto xi + yj + zk$.  Now if $q$ is any nonzero quaternion, then $q$ defines a map $\alpha_q : E \to E$ via $r \mapsto q^{-1}rq$.  This is an isometry because the norm is multiplicative on quaternions.  The map $\alpha_q$ sends 1 to 1, and $E$ is the orthogonal complement of 1, and hence $\alpha_q$ sends $E$ to $E$. Now one can check that $\alpha_q$ is the identity only when $q$ is real.
Edit (from LutzL's comment below): I am taking $\alpha \beta$ to mean do $\alpha$, then $\beta$ (i.e. maps are regarded as right operators).  With the more standard order of left operators ($\alpha \beta$ means do $\beta$ then $\alpha$), you would need $r \mapsto qrq^{-1}$ instead.
