Does every Boolean algebra have an automorphism with exactly zero and one as fixed points? I have been trying to prove off and on for a few weeks that every Boolean algebra has an automorphism that fixes only $0$ and $1$, even the one-element case where $0=1$.
Does every Boolean algebra have such an automorphism? How do you prove it?
For every Boolean algebra that's isomorphic to a powerset algebra, this is obviously true. We can apply a derangement to the singleton sets, and verify by hand that the sole automorphism of the one-element and two-element Boolean algebra has the desired property.
However, there are some weird infinite Boolean algebras out there and I'm having trouble coming up with an argument that works for all of them. There may be a way to use Stone's representation theorem to prove this, but I have zero intuition on how to use it or what having a topological space with clopen sets that look like my original Boolean algebra buys me.
Using an overly complicated argument, I have been able to prove a partial result, namely that, for every Boolean algebra, if there exists an element $c$ that is fixed by every automorphism, then $c$ is $0$ or $1$. At this point I'm stuck, though.

What follows is my attempt to prove a partial result, that if a Boolean algebra has a point $c$ that is fixed by every automorphism, then $c$ is $0$ or $1$.
Let $B_1$ be the sole Boolean algebra where $1=0$, and let $B_2$ and $B_4$ be the $2$ and $4$-element algebras respectively.
Let the light cone of an element $x$ be defined as $\{a : (a \le x) \;\;\text{or}\;\; (x \le a)\}$.
Lemma 101: A Boolean algebra has a fixed point of negation precisely if it is $B_1$.
Let $B$ be a Boolean algebra. Suppose $B$ has a fixed point of negation $c = \lnot c$ .
It follows that $1 = (c \lor \lnot c) = (c \lor c) = c$. However, $1$ is a fixed point of negation if and only if $1 = 0$, i.e. our algebra is $B_1$.
End of proof of Lemma 101.
Theorem 102: If a Boolean algebra has a point $c$ that is fixed by every automorphism, then $c$ is $0$ or $1$.
By inspection, in $B_1$, $B_2$, or $B_4$, $c$ must be $0$ or $1$.
Let $B$ be our Boolean algebra with underlying set $X$ and let $G$ be its automorphism group.
We can partition $X$ into orbits of $G$.
Automorphisms in $G$ cannot reverse comparisons, so we end up with linearly ordered stripes on $X$.
$c$ is in an orbit by itself. By symmetry, $\lnot c$ is also in an orbit by itself.
It therefore follows that the light cone of $c$ and the light cone of $\lnot c$ are the entire algebra $B$.
Here's a picture. Without loss of generality we can make $c$ bigger than $\lnot c$.
   1
  / \
  \ /
   c
  / \
  \ /
  !c
  / \
  \ /
   0

In a Boolean algebra, $c \lor \lnot c = 1$, however, the least upper bound of $c$ and $\lnot c$ is $c$.
Thus $c = 1$.
However, $c$ was fixed to be the larger of $c$ and $\lnot c$, so $c$ is $0$ or $1$ as desired.
End of proof of Theorem 102.
 A: No. Let $B$ be the direct product of the free Boolean algebra on countably many generators (the countable atomless Boolean algebra) with the $2$-element Boolean algebra. In terms of Stone duality this is the Boolean algebra of clopen subsets of the disjoint union $X$ of the Cantor set $\{ 0, 1 \}^{\mathbb{N}}$ and a point $\bullet$. Because $\bullet$ is the unique isolated point of $X$ (equivalently, because $B$ has a unique atom, namely $\bullet$ regarded as a subset of $X$), every automorphism of $X$ fixes this point (equivalently, every automorphism of $B$ fixes its unique atom).
I don't understand your proof of Theorem 2 (which this is a counterexample to), specifically I don't understand what justifies this line:

It therefore follows that the light cone of $c$ and the light cone of $\lnot c$ are the entire algebra $B$.

If you take $c$ to be the unique atom of $B$ above you can see that its light cone does not include any non-empty subset of the Cantor set (regarded as a subset of $X$).
