Suppose $A$ is a complete subalgebra of a complete boolean algebra $B$. Suppose $f : A \to A$ is an automorphism. Then $f$ can be extended to an automorphism of $B$. I can see this using the fact that $B$ is equivalent to a two-step forcing iteration $A * \check B/\dot G$ and using automorphism-of-names. But I am curious, is there a simple and direct way to define this extension of $f$ without reference to forcing?
Edit: Comments show that the premise of my question was wrong. But I learned something so I will leave it to the moderators whether this question needs deleting.
As was brought up in the comments, there is actually a counterexample. I'll give a simpler one than I gave in the comments, with the same idea. Take $B = P(\{1,2,3\})$, and $$A = \{\emptyset,\{1\},\{2,3\},\{1,2,3\}\}.$$ Then $A$ has the automorphism which switches $\{1\}$ and $\{2,3\}$. But this can't extend to an automorphism of $B$, since every automorphism of $B$ must send $\{1\}$ to either $\{2\}$ or $\{3\}$.