Suppose $F$ and $G$ are biadjoint functors, with some fixed unit/counit pairs. How is it that we have an isomorphism of the set of natural transformations $$ \operatorname{Nat}(F,F)\simeq\operatorname{Nat}(G\circ F,1)? $$ I'm used to adjoint functors being characterized by an isomorphism of hom functors like $\hom(F(-),-)\simeq\hom(-,G(-))$, but it's not clear to me how to get from a natural transformation on $F$ to one from $GF$ to the identity.
Say the adjunctions are given by a natural isomorphism $\Phi\colon\hom(F(-),-)\to\hom(-,G(-))$ with unit and counit $\eta\colon 1\to GF$ and $\epsilon \colon FG\to 1$, and $\Phi'\colon\hom(G(-),-)\to\hom(-,F(-))$ with unit and counit $\eta'\colon 1\to FG$ and $\epsilon'\colon GF\to 1$.
$\def\Nat{\textsf{Nat}}$The claim follows from the fact that, if $$ L : {\cal C} \leftrightarrows {\cal D} : R$$ is an adjunction, then $\Nat(LH,K)\cong \Nat(H, RK)$ for all functors $H : {\cal X}\to{\cal C},K : {\cal X}\to {\cal D}$. This, in turn, follows from the fact that $$ \begin{align} \Nat(LH,K) &\cong\int_x {\cal D}(LHx, Kx)\\ &\cong \int_x {\cal C}(Hx, RKx)\\ &\cong\Nat(H, RK). \end{align} $$ Apply this result to the case $G=L, F=R=H, K=1$.