Can non-adjunctions yield equivalences? Whenever $F\dashv U:\mathcal C\leftrightarrow \mathcal D$ with unit and counit $\eta,\varepsilon$, then $F,U$ restrict to an equivalence between the full subcategories with objects $\{A\in\mathcal C\mid\eta_A\, iso\}$ and $\{B\in\mathcal D\mid\varepsilon_B\, iso\}$. (This follows from the triangle identities.)
Is there an example of $F,U$ together with $\eta, \varepsilon$ restricting to an equivalence between $\{A\in\mathcal C\mid\eta_A\, iso\}$ and $\{B\in\mathcal D\mid\varepsilon_B\, iso\}$ such that the tuple $(F,U,\eta, \varepsilon)$ isn't an adjunction?
 A: 
This follows from the triangle identities

The triangle identities aren't necessary if $\varepsilon_x$ and $\eta_y$ are always isomorphisms. In that case, $F$ and $U$ are inverses of each other without restricting. That's just because the data of being an inverse just requires natural isomorphisms $FU \cong 1$ and $UF \cong 1$, with no additional requirements (like the triangle identities).
So to get an example of an equivalence which is not adjoint, we just need to arrange for $\varepsilon$ and $\eta$ to be isomorphisms, but for the triangle identities to not hold.
One possibility is to use the category with one object and one non-identity isomorphism $f$ from that object to itself (as well as $f^2$, $f^3$... and $f^{-1}$, $f^{-2}$...). This is the delooping of the group of integers, so if you want to think of the morphisms as integers, composition is just addition.
Let $F$ and $U$ both be the identity functor on this category. Let $\varepsilon_x$ be the identity from $x$ to itself (remember that $FUx = x$) and let $\eta_y = f$.
This choice of $\eta$ is natural because every morphism in this category is some power of $f$, so it commutes with $\eta$. Let $g \colon y \to y'$ be some morphism. Then $g = f^k$ for some integer $k$ (possibly negative).
$$
\begin{align}
\eta_{y'} \circ g &= f \circ f^k \\
&= f^{k + 1} \\
&= f^k \circ f \\
&= g \circ \eta_y \\
&= UF(g) \circ \eta_y
\end{align}
$$
Since the identity is always a natural isomorphism and $f$ is assumed to be an isomorphism, we have the data needed for an equivalence.
However, $U \varepsilon_x \circ \eta_{U x} = 1 \circ f = f \ne 1$. Additionally, $\varepsilon_{F y} \circ F \eta_y = 1 \circ F(f) = f \ne 1$. So the triangle identities don't hold and $(F, U, \eta, \varepsilon)$ is not an adjunction.
The problem here lies entirely with $\eta$ and $\varepsilon$. In fact, it's always possible to improve an equivalence to an adjoint equivalence by choosing $\varepsilon$ correctly (Theorem 3.3). Here, if we made $\varepsilon_x = f^{-1}$, everything would work out again.
