Don Monk's answer confirms my expectation; let me now also provide a more precise statement and a proof:
If $A$ is a Boolean algebra with a trivial automorphism group, then the automorphism group of $A\times A$ is canonically isomorphic to $(A,+)$, acting by $a\cdot(b,c)=((1-a)b+ac,(1-a)c+ab)$.
By Stone duality, this amounts to proving: if $X$ is a Stone space (totally disconnected, Hausdorff compact topological space) with trivial self-homeomorphism group, then any self-homeomorphism of $X\times\{-1,1\}$ is given by $(x,t)\mapsto (x,u(x)t)$ where $u$ is a continuous function $X\to\{-1,1\}$.
To prove the latter, let us first check that for any two clopen subsets $Y,Z$ of $X$ and homeomorphism $h:Y\to Z$, we have $Y=Z$ and $h$ is the identity. Indeed, otherwise there exists $y\in Y$ such that $h(y)\neq y$. By passing to smaller clopen subsets, we can then assume that $Y$ and $Z$ are disjoint. Then we can extend $h$ to a self-homeomorphism of $X$, as equal to $h^{-1}$ on $Z$ and identity elsewhere. (Note that in this part, we use that $X$ is Hausdorff with a basis of clopen subsets, but compactness does not play any role).
Then it's easy to conclude. Let $h$ be a self-homeomorphism of $X\times\{-1,1\}$. Call "component" the two subsets $X\times\{t\}$. We find a finite clopen partition of $X$ such that for each part $Y$ of the partition, each of $Y\times\{t\}$ is mapped into a single component. Then by the previous fact on partial homeomorphism, we have $h(y,t)=(y,s(y,t))$ for some $s(y,t)\in\{-1,1\}$ (which is constant when $y$ varies in $Y$). By injectivity, we have $s(y,1)=-s(y,-1)$. So we can write $s(y,t)=u(y)t$. $\Box$
Similarly, for a finite set $F$, the automorphism group of $X\times F$ consists of the $(x,t)\mapsto (x,\sigma(x)(t))$, where $\sigma$ ranges over the continuous functions $X\to\mathfrak{S}(F)$. The corresponding Boolean algebra action can be described accordingly.
