Let's start with a functor $U:{\mathcal A}\to{\mathcal X}$ that's full and faithful. Bypassing questions of Choice, it's also essentially surjective on objects in the sense that there is a function $F:{\mathsf ob}{\mathcal X}\to{\mathsf ob}{\mathcal A}$ along with an assignment of an isomporhism $\eta_X:X\to U F X$ to each object $X$.
From these data, for each morphism $f:X\to Y$ of $\mathcal X$, we define $F f:F X\to F Y$ as the unique $\mathcal A$-map such that
$$ \eta_X^{-1};f;\eta_Y = U(F f), $$
in other words such that $\eta$ is natural. It is easy to check that $F$ preserves identity and composition.
It remains to define $\epsilon$. For one of the triangle laws we require $U\epsilon_A=\eta_{U A}^{-1}$, which uniquely defines $\epsilon_A$ since $U$ is full and faithful. Naturality of $\epsilon$ follows from that of $\eta$.
The other triangle law is $\eta_{U A};U\epsilon_A={\mathsf id}$, for which it suffices that this hold with $U$ applied, since that's full and faithful.
By naturality of $\eta$ and the first triangle law, we have
$$ \eta_X;U F\eta_X; U\epsilon_{F X} = \eta_X;\eta_{U F X}; U\epsilon_{F X} = \eta_X $$
but $\eta_X$ is invertible an $U$ is full and faithful, so the other triangle law holds.
In other words, the obvious data for "full, faithful and essentially surjective on objects" yield an adjoint equivalence.
So what other kind of equivalence is there? If the isomorphism $\eta$ in "essential surjectivity" is to be natural, it can only be as above. However, the other isomorphism $\epsilon'$ could come from somewhere else. Nevertheless, $\eta_{U A};U\epsilon'_A$ is still a natural automorphism of $U A$, which must be $U$ applied to a natural isomorphism of $A$.
In other words, a non-adjoint equivalence is given by arbitrary natural isomorphisms applied to an adjoint equivalence.
This is a situation that can easily be realised with group isomorphisms, yielding the counterexample that was requested.
