Adjunctions restrict to an equivalence From this question:

Let a pair of functors $F : \mathscr{A} \rightarrow \mathscr{B}$ and $G : \mathscr{B} \rightarrow \mathscr{A}$ be adjunction such that $F$ is left adjoint to $G$, i.e. $F \dashv G $. Write $\textbf{Fix}(GF)$ for the full subcategory of $\mathscr{A}$ whose objects are those $A \in \mathscr{A}$ such that $\eta_{A}$ is an isomorphism, and dually $\textbf{Fix}(FG) \subseteq \mathscr{B}$. Prove that the adjunction $(F, G, \eta, \epsilon)$ restricts to an equivalence $(F', G', \eta', \epsilon')$ between $\textbf{Fix}(GF)$ and $\textbf{Fix}(FG)$.

So we need to prove that the functor $F':\textbf{Fix}(FG)\to \textbf{Fix}(FG)$ defined by $F'(A)=F(A),\ F'(f)=F(f)$ is full, faithful, and essentially surjective on objects.
Essential surjectivity on objects seems to be obvious: if $B\in \textbf{Fix}(FG)$, then since $\epsilon_B:FG(B)\to B$ is an isomorphism, $B$ is isomorphic to $F(A)$ where $A=G(B)$.
But fullness and faithfulness is not as obvious to me. 
Fullness: Suppose there is an arrow $F(A)\to F(A')$ for some $A,A'\in \mathscr A$. We need to prove that this arrow is the image (under $F$) of some arrow $A\to A'$ in $\mathscr A$. Since $F\dashv G$, there is a bijection $$\mathscr B (F(A),F(A'))\cong\mathscr A(A,GF(A'))$$
such that some naturality conditions hold. (And $GF(A')$ is isomorphic to $A'$ via $\eta_A$.) But the bijective correspondence from right to left is not necessarily given by $F(-)$. I don't have other ideas how to find the preimage of $F(A)\to F(A')$.
Faithfulness: Suppose there are two arrows $f,g:A\to A'$ such that $F(f)=F(g):F(A)\to F(A')$. We need to show that $f=g$. I can't think of any applicable technique of proving that two arrows are equal. Nor do I understand how to use that $F(f)=F(g)$.
 A: Let $$F': \mathbf{Fix}(GF)\to \mathbf{Fix}(FG)$$ be the functor defined by $F'(A)=F(A)$ and $F'(f)=F(f)$ for all objects $A$ and arrows $f$ in $\mathbf{Fix}(GF)$. (See this question for a proof of the fact that $F'(A)\in\mathbf{Fix}(FG)$ and that $F'(f)$ is a morphism in $\mathbf{Fix}(FG)$.) Similarly, let $$G':\mathbf{Fix}(FG)\to\mathbf{Fix}(GF)$$ be the functor defined by $G'(B)=G(B)$ and $G'(g)=G(g)$ for all objects $B$ and arrows $g$ in $\mathbf{Fix}(FG)$.
Let $\eta':1_{\mathbf{Fix}(GF)}\to G'\circ F'$ be the natural transformation whose component at $A\in \mathbf{Fix}(GF)$ is given by $\eta'_A=\eta_A$. This is indeed a natural transformation because $\eta$ is. And let $\epsilon':F'\circ G'\to 1_{\mathbf{Fix}(FG)}$ be the natural transformation whose component at $B\in \mathbf{Fix}(FG)$ is $\epsilon_B'=\epsilon_B$.
Now it is almost immediate that $(F',G',\eta',\epsilon')$ is an equivalence. To establish this, we need to prove that $\eta'$ and $\epsilon'$ are natural isomorphisms, but they indeed are because their components are isomorphisms. (If $A\in \mathbf{Fix}(GF)$, then by the definition of $\mathbf{Fix}(GF)$, $\eta_A:A\to GF(A)$ is an isomorphism, hence so is $\eta'_A:A\to GF(A)$; and similarly for $\epsilon$.)
