When are Categorical Properties of Maps Preserved Under An Adjoint Isomorphism? Suppose that we a functor $F : \mathcal{A} \to \mathcal{B}$. Then we say that a functor $G : \mathcal{B} \to \mathcal{A}$ is right-adjoint to $F$ if $\operatorname{Hom}_{\mathcal{B}}(F(A),B) \cong \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ for all objects $A \in \mathcal{A}$, $B \in \mathcal{B}$.
I am interested in whether the "adjoint isomorphism" $\operatorname{Hom}_{\mathcal{B}}(F(A),B) \cong \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ necessarily preserves any categorical properties of the maps in either Hom set.
That is to say, if we explicitly label the adjoint isomorphism
$$
\vartheta : \operatorname{Hom}_{\mathcal{B}}(F(A),B) \overset{\cong}{\to}\operatorname{Hom}_{\mathcal{A}}(A,G(B)),
$$
then can we say if, or when it is the case, that a map $(F(A) \overset{T}{\to} B) \in \operatorname{Hom}_{\mathcal{B}}(F(A),B)$ has categorical property X if and only if $(A \overset{\vartheta(T)}{\to} G(B)) \in \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ has the same categorical property X?
By this question, we know that the adjoint isomorphism does not neccesarily preserve epimorphisms, so we can exclude this from being a candidate property X. However are there any other properties that are conserved?
Failing this, can we impose extra conditions (exactness?) on $F$ and $G$ that would allow certain properties to be conserved?

For context, I am typically concerned with functors between module categories, and in particular induction / restriction (or similar adjoint pairs).
 A: (Note that in the definition, there should be a "naturally in $A,B$")
The answer is that usually they don't, and the context of induction/restriction adjunctions is the perfect place to put all of these to the test and see that they mostly fail.
For instance say you have a ring map $A\to B$ and are looking at extension/restriction of scalars; I will let $U$ denote the forgetful functor.
Then

$f$ can be an isomorphism without its adjoint being one.

Example : take $A\to UB$ : this is clearly not an isomorphism in general (it is one if and only if $A\to B$ is one); however its adjoint is $B\cong B\otimes_A A\to B$ which is an isomorphism.
Dually, if you take $UB\to UB$, this is an isomorphism of $A$-modules, but its adjoint $B\otimes_A B\to B$ is not necessarily one (it will be if $A\to B$ is a localization or a quotient for instance, but not for general ring maps)
As you can see, there's no reasonable general condition you can impose on $A\to B$ for this to hold, and "$f$ is an isomorphism" is one of the most basic categorical property.
As pointed out in the comments, in general (if $F$ is the left adjoint) the adjoint of $f: FM\to N$ is given by the composite $M\to UFM\to UN$, and so it factors through $\eta_M$. So the properties you are interested in will have to be behaved well with respect to $\eta_M$, which does not depend on $f$ and its categorical properties, which explains why many of these things fail : $\eta$ typically does not have these properties (epimorphism, monomorphism, isomorphism, etc.)
(and dually for $\epsilon$)
So you'll want to look for properties that $\eta$ has,that are preserved by $G$ and stable under composition (or dually, that $\epsilon$ has, that are preserved by $F$ and stable under composition), and these are rare without stringent restrictions on your adjunction (e.g. one of the functors is full/faithful/fully faithful which is somewhat rare)
This isn't a full answer because I can't say "there are none", but ultimately it depends on what restrictions you're ready to impose on your functors. If you want to stay reasonably general, there are basically none that will hold.
There is an important counterexample to what I just said, though ! Suppose $\mathcal{A,B}$ are pointed, that is, they have a zero object. Then it is preserved by the left and the right adjoint, and so we get that a morphism $f$ is $0$ (i.e. factors through the zero object) if and only if its adjoint does.
In the case of additive categories, this can also be seen by the fact that $\hom(F(A),B)\cong \hom(A,G(B))$ is automatically an isomorphism of abelian groups; and so preserves the $0$ morphism, but it is true in the more general pointed case.
