New discovery of the unconventional matrix representation for the quaternion $H_8$ Let us start with the notation for the quaternion group  of order 8 as $H_8=\{e,i,j,k,g,i^{-1},j^{-1},k^{-1}\}$ where $ij=k=j^{-1}i$, $i^2=j^2=k^2=g$ and $i^4=j^4=k^4=g^2=e$. 
One way to write the 4-by-4 matrix representation of the quaternion as $H_8$ is that
$e=\begin{bmatrix}
 1 & 0 & 0 & 0 \\ 
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 
\end{bmatrix}, 
i=\begin{bmatrix}
 0 & -1 & 0 & 0 \\ 
 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -1 \\
 0 & 0 & 1 & 0 
\end{bmatrix},
j=\begin{bmatrix}
 0 & 0 & -1 & 0 \\ 
 0 & 0 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0 
\end{bmatrix},
k=\begin{bmatrix}
 0 & 0 & 0 & -1 \\ 
 0 & 0 & -1 & 0 \\
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 
\end{bmatrix}.$
$$g=i^2=j^2=k^2=-\begin{bmatrix}
 1 & 0 & 0 & 0 \\ 
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 
\end{bmatrix}=-e.$$
And we notice that the matrices of $i,j,k$ are skew symmetric. See for example the Ref: Matrix representation of quaternions
Farebrother-Groß-Troschke and please check the Wikipedia.

Now let us consider the new notation for the quaternion group  of order 8 as $H_8=\{e,i',j',k',g',i'^{-1},j'^{-1},k'^{-1}\}$. My question is that can we choose the non-skew symmetric matrix representation such that we design and fix the new $$k'=\begin{bmatrix}
 0 & 0 & 0 & -1 \\ 
 0 & 0 & 1 & 0 \\
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 
\end{bmatrix}$$ to be rather unconventional, and notice that the $g'=k'^2$ has a rather unconventional definition
  $$g'=k'^2=
\begin{bmatrix}
 -1 & 0 & 0 & 0 \\ 
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & -1 
\end{bmatrix},$$ but the identity $$e=(g')^2=i'^4=j'^4=k'^4=\begin{bmatrix}
 1 & 0 & 0 & 0 \\ 
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 
\end{bmatrix}$$ is still the conventional identity matrix $e$.
The questions are that does there exist such non-skew symmetric matrix representation of 4-by-4? For the real matrix or the complex matrix?  Given that we restrict the $k'=\begin{bmatrix}
 0 & 0 & 0 & -1 \\ 
 0 & 0 & 1 & 0 \\
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 
\end{bmatrix}$, what will be the matrix representation form of $i',j'$? (The real matrix or the complex matrix?)

 A: Let $v_1,v_2,v_3,v_4$ be a basis for such a representation. We see that $k'$ acts with eigenvalues $\pm1$ on the span $V$ of $v_2,v_3$, and with eigenvalues $\pm i$ on the span $W$ of $v_1,v_4$. This suggests that we should be able to build at least a complex representation with $k'$. 
On $V$ we can simply let $j,j^{-1},k,k^{-1}$ all act the same way, and let the subgroup $K=\{1,i,g,i^{-1}\}$ act as the identity. This is the trivial representation of $K$ induced to all of $H_8$, and we only need real matrices to do that.
On $W$ I think we need complex matrices as the resulting representation must be the usual 2-dimensional representation of $H_8$. Putting all these pieces together we end up with
$$
i'=\left(\begin{array}{cccc}
\bf{i}&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&-\bf{i}\end{array}\right),
\quad
j'=k'i'=\left(\begin{array}{cccc}
0&0&0&\bf{i}\\
0&0&1&0\\
0&1&0&0\\
\bf{i}&0&0&0
\end{array}\right).
$$
In the matrices above I used boldface $\bf{i}$ for a complex square root of $-1$. Just to make sure it won't get confused with the quaternion $i\in H_8$ - admittedly there should be no confusion here (but this issue would appear when jotting down elements of the complex group algebra $\Bbb{C}H_8$).
We could equally well induce from $K'=\{1,j,g,j^{-1}\}$ instead of $K$. This would make the matrices $i'$ and $j'$ trade their central 2x2 block (their action on the subspace $V$).

Another way of seeing why the 2x2-block giving the action on $W$ (=the corners of the 4x4-matrices) cannot be all reals is that the real span of those matrices must give an algebra isomorphic to the Hamiltonian quaternions. But that algebra cannot be embedded into 2x2 real matrices, because it has real dimension four, hence would have to be all of $M_2(\Bbb{R})$. But that matrix algebra has zero divisors unlike the quaternions, so this is a contradiction.
