For two matrices to commute, it is necessary that each matrix preserves the eigenspace of the other matrix (that is, it can't map part of an eigenspace onto a different eigenspace). As multiples of the identity matrix do not change eigenspaces and have all vectors in the same eigenspace, they commute with all other matrices.
If $A$ and $B$ commute, and $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, then,
ABx=BAx=B\lambda x=\lambda Bx
and thus $Bx$ must also be an eigenvector of $A$ with the same eigenvalue... or a zero vector.
Suppose that $v=(a,b,c,d)$ is an eigenvector of our matrix. Then we must find, in the same eigenspace, $(a,b,-c,-d)$, $(d,c,-b,-a)$, $(-d,c,b,-a)$, and $(c,-d,-a,b)$ - (I have dropped the $i$ from the third matrix, as it's just a constant multiplier).
Adding and subtracting the middle two together, we can also see that $(0,c,0,-a)$ must be in the eigenspace, as must $(d,0,-b,0)$. At least one of these must be non-zero. Let's assume that $(0,c,0,-a)$ is non-zero.
Then we can also see that $(0,c,0,a)$ is in the eigenspace (from the first matrix), and so both $(0,c,0,0)$ and $(0,0,0,a)$ are in the eigenspace. Let's assume that $(0,0,0,a)$ is non-zero, and thus $(0,0,0,1)$ is in the eigenspace. Now, from the last matrix, $(0,0,1,0)$ is in the eigenspace. From the third matrix, we can then determine that $(0,1,0,0)$ and $(0,0,0,1)$ are in the eigenspace.
A similar analysis works if you assume $b$, $c$, or $d$ is non-zero.
Therefore, the eigenspace is the set of all 4D vectors, all sharing the same eigenvalue. This tells us that our matrix must be the identity matrix multiplied by that eigenvalue.