$\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}$This is a quick question - my proof is very short, but I’m suspicious of it.
Suppose the categories $\A,\B$ are equivalent via the functors $F:\A\to\B,\,G:\B\to\A$ and natural isomorphisms $\varphi:FG\to1_{\B}\,\psi:1_{\A}\to GF$. We want to show that $F\dashv G$. To that end, it is simplest to consider terminal morphisms. Let $\varepsilon_B:=\varphi_B$ for every $B\in\B$. Then I claim $(G(B),\varepsilon_B)$ is always a terminal morphism from $F$ to $B$.
Suppose $p:F(A)\to B$ in $\B$ for an arbitrary $A\in\A$. Consider the arrow $h:=\varphi_B^{-1}\circ p$: since $F$ is an equivalence of categories, it is full and faithful, therefore $h:F(A)\to FG(B)$ implies the existence of a unique $q:A\to G(B)$ such that $F(q)=h$. But then for all such $p$, there exists a unique $q$ such that $\varepsilon_B\circ F(q)=p$, so $(G(B),\varepsilon_B)$ is a terminal morphism from $F$ to $B$, so $F\dashv G$. $\blacksquare$
Why am I suspicious of this proof? Because I have been told from many sources that the isomorphisms that make $F,G$ equivalences are not the unit and counit of the adjunction, in general, but I have here used one of them as the counit. Have I done something wrong, or is this a correct proof?