Are epimorphisms (defined via an obvious action) of free Boolean algebras whose set of generators is a group automorphisms? Let $G$  be a group. Consider $B$, the free Boolean algebra with generating set (I'll call them "variables") $G$. Let $F$  be some formula (that is, some fixed element of $B$). Define an endomorphism ϕ  on $B$  by sending the identity element of $G$  to $F$, and then sending an element $g$  of $G$  to the formula you get from $F$  by appropriately changing the variables in $F$. So for example, if $F$  is $h$∨∼$k$  then send $g$  to $(gh)$∨∼($gk$) . 
Now assume that ϕ  is onto. Is ϕ  an automorphism?
 A: The answer is yes if every finitely generated subgroup of $G$ is residually finite.  First, note that if $G$ is finite, this is automatic, since a surjection from a finite set to itself is automatically injective.  For general $G$, note that $\phi$ is surjective iff there is a generator of $B$ in the image of $\phi$ (because then every other generator will also be in the image by the translation invariance of $\phi$).
Now suppose $\phi$ is surjective but not injective; say $\phi(x)=\phi(y)$ for two distinct formulas $x$ and $y$.  Then there are finitely many elements of $G$ that collectively witness the fact that $\phi$ is surjective but not injective: take the finitely many variables appearing in $F$, the finitely many variables appearing in some formula that $\phi$ sends to a generator, and the finitely many variables appearing in $x$ and $y$.  If every finitely generated subgroup of $G$ is residually finite, then the subgroup generated by these elements has a finite quotient $H$ that still recognizes that $x$ and $y$ are distinct.  The image of the formula $F$ in $H$ will define an endomorphism of the free Boolean algebra on $H$ which is surjective but not injective, contradicting the finiteness of $H$.
