Let $$R=A_1\times A_2\times\dots\times A_n.$$ Let's assume there are no non trivial automorphisms of these $A_i$s. I claim that the only possible automorphisms of this structure are the permutations of the $A_i$s. So, there are a total of $n!$ automorphism possible over this structure.
Question is how to prove this. I am proceeding in this way: let $R\cong A\times (0)$, $(0)$ being the zero ring. Now, $(a,0)$, $a\in A$, can only get mapped to $(a,0)$ or $(0,a)$. This can be established via some simple isomorphism argument. But I am not been able to generalize this to any ring product. Any help are welcome.
You meant that there are at most $n!$ possible automorphisms all given by permuting the isomorphic $A_j$.
RghtHndSd showed a counter-example, eg. $A_1=A_2=\Bbb{F}_p\times \Bbb{Q}$ then $((a,b),(c,d))\to ((a,d),(c,b))$ is an automorphism of $A_1\times A_2$.
If the $A_j$ are integral domains then your claim is true, because the $e_j = (1,\ldots,\underbrace{0}_{j \ th},1,\ldots)$ are the only elements such that $e_j R$ is a prime ideal and $e_j^2=e_j$. An automorphism $\sigma\in Aut(R)$ must permute them. So it permutes the $1-e_j$, thus it induces an isomorphism $$A_j\cong (1-e_j) R \quad to\quad A_{\sigma_j} \cong (1-e_{\sigma_j}) R$$ And two distinct isomorphisms $A_j\to A_{\sigma_j}$ give a non-trivial automorphism of $A_j$, that you assume to not exist. Thus $\sigma$ is determined by the $\sigma_j$.