Eigenvalues of Group Elements and Quaternions The quaternion algebra is given by
$\mathbb{H}$ = $\{a+bi+cj+dk \mid a, b, c, d \in \mathbb{R}, i^2 = j^2 = k^2 = -1, ij = k = -ji\} := \{z_1+z_2j \mid z_1, z_2 \in \mathbb{C}\}.$
I consider the 3-sphere as the set of unit quaternions (the quaternions of length 1) as follows:
$\mathbb{S}^3 = \{a + bi + cj + dk \mid a^2 + b^2 + c^2 + d^2 = 1\} = \{z_1 + z_2j \mid |z_1|^2 + |z_2|^2 = 1\}$
The product in $\mathbb{H}$ induces a group structure on $\mathbb{S}^3$. For each pair $(p, q)$ of elements of $\mathbb{S}^3$, the function 
$\Phi_{p,q} : \mathbb{H} \rightarrow \mathbb{H}$ 
with $\Phi_{p,q}(h) = phq^{-1}$ leaves invariant the length of quaternions. 
We can, therefore, define a homomorphism of groups:
$\Phi : \mathbb{S}^3 \times \mathbb{S}^3 \rightarrow \mathrm{SO}(4)$ such that $\Phi(p, q) = \Phi_{p,q}$.
I am looking at subgroup $C_4 \times D^*_{24}$ of  $\mathbb{S}^3 \times \mathbb{S}^3$, where $C_4$ is the cyclic group of order 4 and $D^*_{12}$ is the binary dihedral group of order 12. I have looked at 3 different ways of representing this group and in each case I get resulting matrices in $\mathrm{SO}(4)$ with different eigenvalues. Here are the 3 ways:


*

*$C_4 = \{\langle g \rangle$, where $g^4 = 1\}$, $D^*_{12} = \{\langle A, B \rangle$, where $A^{6} = 1, B^4 = 1$ and $BAB^{-1} = A^{-1}\}$.
I represent $g$ as the matrix 
\begin{pmatrix}
 i & 0 \\
 0 & -i 
\end{pmatrix}
I represent $A$ as the matrix 
\begin{pmatrix} 
 e^{i\pi/3} & 0 \\                   
 0 & e^{-i\pi/3} \end{pmatrix}. 
I represent $B$ as the matrix 
\begin{pmatrix} 
 0 & i \\                 
 i & 0 \end{pmatrix}. 
With this, $BA$ is the matrix 
\begin{pmatrix} 
0 & e^{i\pi/6} \\                    
-e^{-i\pi/6} & 0 \end{pmatrix}. 
Then $\Phi(g, BA)(h)$ is the $4 \times 4$ matrix 
\begin{pmatrix} 
0 & 0 & \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \\\\                    
0 & 0 & -\cos \frac{\pi}{6} & \sin \frac{\pi}{6} \\\\
\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0 & 0 \\\\                    
-\cos \frac{\pi}{6} & \sin \frac{\pi}{6} & 0 & 0 \end{pmatrix}. 
This has eigenvalues $\pm e^{\pm i\pi/3}$


*I  can also view $D^*_{12} = C_{6} \cup C_{6}j$


With $g$ and $A$ as before, I represent $j$ as the matrix 
\begin{pmatrix} 
 0 & 1 \\                    
-1 & 0 \end{pmatrix}. 
With this, $Aj$ is the matrix 
\begin{pmatrix} 
\ 0 & e^{i\pi/3} \\                   
-e^{-i\pi/3} & 0 \end{pmatrix}. 
Now, $\Phi(g, Aj)(h)$ is the $4 \times 4$ matrix 
\begin{pmatrix} 
0 & 0 & \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \\\\                    
0 & 0 & -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\\\
\sin \frac{\pi}{3} & \cos \frac{\pi}{3} & 0 & 0 \\\\                    
-\cos \frac{\pi}{3} & \sin \frac{\pi}{3} & 0 & 0 \end{pmatrix}. 
This has eigenvalues $\pm e^{\pm i\pi/6}$


*I represent $g = i$, $A = e^{i\pi/3}$ and $j = j$.
With this 
$\Phi(i, Aj)(h)$ = $i (a + bi + cj + dk) (-jcos(\pi/3) - ksin(\pi/3))$
= $c\sin(\pi/3)-d\cos(\pi/3) + i(c\cos(\pi/3) + d\sin(\pi/3)) + j(a\sin(\pi/3)+ b\cos(\pi/3)) + k(-a\cos(\pi/3) + b\sin(\pi/3))$ 
which gives the $4 \times 4$ matrix for $\Phi(i, Aj)$ to be
\begin{pmatrix} 
0 & 0 & \sin \frac{\pi}{3} & -\cos \frac{\pi}{3} \\\\                    
0 & 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\\\
\sin \frac{\pi}{3} & \cos \frac{\pi}{3} & 0 & 0 \\\\                    
-\cos \frac{\pi}{3} & \sin \frac{\pi}{3} & 0 & 0 \end{pmatrix}. 
This has eigenvalues $\pm 1, \pm 1$
So I have 3 different ways to do this and in each case, I get a different matrix in $\mathrm{SO}(4)$ with different eigenvalues. I thought that if the three matrices represent the same element in $\mathrm{SO}(4)$, their eigenvalues should have been the same. What mistake am I making here?
 A: This answer is an expanded version of my comment above.  In short: your third computation is correct, your second one has a mistake, and your first one computes a different thing.
First, let's make some things more precise.  There are several subgroups of $\mathbb{S}^3\times \mathbb{S}^3$ which are isomorphic to $C_4\times D_{12}^*$, so we must define the one we wish to study.  Judging from your point 3., you seem to want $C_4$ to be generated by $i$, and $D_{12}^*$ to be generated by $e^{i\pi/3}$ and $j$.
Now, you want to represent quaternions as $2\times 2$ matrices with values in $\mathbb{C}$.  Note that there are several ways to do so (compare the Wikipedia pages in English and in French, for instance).  The one you seem to be using sends the quaternion $a+bi+cj+dk$ to the matrix $\pmatrix{a+bi & c+di \\ -c+di & a-bi}$.
Let's compute $\Phi(i,Aj)$.  Your computation in point 3. is exactly right, so I won't add anything about it in this answer.
For your point 2., you represent $i$ as the matrix $\pmatrix{i & 0 \\ 0 & -i}$ and $Aj$ as the matrix $\pmatrix{0 & e^{i\pi/3} \\ -e^{-i\pi/3} & 0}$, which is correct.  This means that $(Aj)^{-1}$ is represented by the matrix $\pmatrix{0 & -e^{i\pi/3} \\ e^{-i\pi/3} & 0}$.  Thus, if $\pmatrix{a+bi & c+di \\ -c+di & a-bi}$ represents an arbitrary quaternion, the map $\Phi(i,Aj)$ acts on it in the following way (in what follows, $\sin$ and $\cos$ stand for $\sin(\pi/3)$ and $\cos(\pi/3)$, respectively):
$$ \pmatrix{i & 0 \\ 0 & -i}\pmatrix{a+bi & c+di \\ -c+di & a-bi}\pmatrix{0 & -e^{i\pi/3} \\ e^{-i\pi/3} & 0} = \pmatrix{c\sin - d\cos +i(c\cos +d\sin) & a\sin + b\cos +i(-a\cos + b\sin) \\ -a\sin -b\cos +i(-a\cos +b\sin) & c\sin -d\cos -i(c\cos +d\sin)}.  $$
This means that the matrix of $\Phi(i,Aj)$ is
$$ \pmatrix{0 & 0 & \sin(\pi/3) & -\cos(\pi/3) \\ 0 & 0 & \cos(\pi/3) & \sin(\pi/3) \\ \sin(\pi/3) & \cos(\pi/3) & 0 & 0 \\ -\cos(\pi/3) & \sin(\pi/3) & 0 & 0}.$$
This matrix is the same as in your point 3.; I suspect you have made a mistake when computing this matrix in your point 2., and that this is why the matrices differ.
Finally, for your point 1., note that the element represented by your matrix $B$ is the quaternion $k$.  Thus $BA$ represents the quaternion $ke^{i\pi/3} = k\cos(\pi/3)+j\sin(\pi/3)$, which is different from the element $e^{i\pi/3}j = j\cos(\pi/3) + k\sin(\pi/3)$ used in points 2. and 3.  This explains why the matrix in your first calculation is different.
