Please let me know if you think this proof is good enough or if some misconceptions are present.
$G$ is a non-trivial group. Define elements of Aut($G\times G \times G)$ as follows:
$\phi_1[(g,h,k)]=(g,h,k)$ (identity)
$\phi_2[(g,h,k)]=(g,k,h)$
$\phi_3[(g,h,k)]=(k, g,h)$
$\phi_4[(g,h,k)]=(k,h, g)$
$\phi_5[(g,h,k)]=(h,k, g)$
$\phi_6[(g,h,k)]=(h,g, k)$
Define a set $S:=\{\phi_j:1\le j\le 6\}$.
Subgroup tests:
-> $S$ is clearly not empty.
-> $\phi_i \phi_j^{-1}(x)=[\phi_i(\phi_j(x))]^{-1}=\phi_j(\phi_i(x))=\phi_j\phi_i(x)$. Hence, $S$ is a subgroup.
(I'm wondering if the above is rigorous enough to prove that $S$ is a subgroup).
Isomorphism:
Define $\Phi: S\to S_3$ in such a way that $\phi_1$ goes to $(1)$, $\phi_2$ goes to $(23)$, $\phi_3$ goes to $(132)$, $\phi_4$ is sent to $(13)$, $\phi_5$ is sent to $(123)$, and $\phi_6$ is sent to $(12)$.
We can thus see that $S<$ Aut($G\times G \times G$) and $S \cong S_3$.
There might be a shorter way to prove the above.
For subgroup test take $x=(g,h,k)$, and then show $\phi_i \phi_j^{-1} (x) \in S$.
Then you can check $S$ is non-abelian group of order 6, and only non-abelian group of order 6 is $S_3$.