In this entry I derive the fact $\Aut(Q_8) \cong S_4$, making use of some beautiful observations.
Recall that $Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ is generated by the elements $-1$, $i$, $j$ and $k$ under the relations
$$(-1)^2=1 \quad\text{and}\quad i^2=j^2=k^2=ijk=-1.$$
This is not a minimal set of generators and relations, but it is the most appropriate choice for my argument. Now let $\varphi\in \Aut(Q_8)$. Straight away, we can already deduce that $\varphi$ fixes $1$ and $-1$, since they are the only order $1$ and $2$ elements respectively. Hence, $\varphi$ can be thought of as a permutation of the remaining 6 elements $\{\pm i,\pm j,\pm k\}$.
There are restrictions on what permutations are allowed. The first set of restrictions is
$$\begin{align*} \varphi(-i) &= -\varphi(i) \\ \varphi(-j) &= -\varphi(j) \\ \varphi(-k) &= -\varphi(k). \end{align*}$$
This can be formalised as follows: $\varphi$ preserves the grouping of $\{i,-i\}$, $\{j,-j\}$ and $\{k,-k\}$---that is, $\varphi$ sends each set into another one of those sets. There are no restrictions on the permutations of the sets, as well as the permutation of elements within each set, so the subgroup of $S_6$ preserving grouping is the wreath product
$$S_2 \wr S_3 \cong (S_2 \times S_2 \times S_2) \rtimes S_3,$$
which has $2^3\cdot6 = 48$ elements.
Still, not every permutation in the wreath product is allowed. Consider the case where $\varphi$ maps $i \mapsto j$, $j \mapsto i$ and $k \mapsto k$. It can be seen that
$$\varphi(ij) = k \neq -k = \varphi(i)\varphi(j),$$
so $\varphi$ is not actually an automorphism. In general, the second restriction comes from the relation $ijk=-1$: applying $\varphi$ to both sides yields
$$\begin{equation} \label{eq:varphi-relation} \varphi(i)\varphi(j)\varphi(k) = -1. \end{equation}$$
This is where I had a brilliant idea. I imagined that the product $ijk$ represented the identity matrix $I = \begin{pmatrix}\hat\i & \hat\j & \hat k\end{pmatrix}$; then, I interpreted the statement $ijk=-1$ as saying that $I$ has negative orientation. Following through with this idea and \eqref{eq:varphi-relation}, $\Aut(Q_8)$ can now be thought of as the group of 'orientation-preserving' permutations! The idea actually makes sense: the permutation swapping $i$ and $j$ is 'orientation-reversing', and so is the permutation sending $i$ to $-i$. On the other hand, the permutation mapping $i \mapsto j$, $j \mapsto k$ and $k \mapsto i$ is 'orientation-preserving'.
This idea can be made precise: recall that each automorphism $\varphi$ is determined by its action on $i$, $j$ and $k$. We can then associate $\varphi$ with a linear isomorphism of $\R^3$ by replacing those elements as basis vectors. Specifically, $\varphi$ is a signed permutation matrix: a permutation matrix whose nonzero entries can be $\pm1$.
As shown above, the group $G$ of signed permutation matrices has order $48$ and is isomorphic to $S_2 \wr S_3$. Then, $\Aut(Q_8)$ is the kernel of the determinant homomorphism from $G$ to $C_2$, consisting of the orientation-preserving matrices. Hence, we have $|\Aut(Q_8)| = |G|/|C_2| = 24$. Finally, it is easily seen that the subgroup of orientation-preserving matrices is isomorphic to the symmetry group of the cube, proving $\Aut(Q_8) \cong S_4$.
It is very tempting to think that if higher-dimensional analogues of $Q_8$ exist, then the preceding argument can be easily generalised. For example, is there a version of $Q_8$ but with the elements $i$, $j$, $k$ and $l$?