What is a natural bijection? Here is the definition of adjoint functors taken from here:

and here is a property about universal arrows:

To me the concept of natural bijection is imprecise. Does it coincide with the notion of natural isomorphism (so a natural transformation that has an inverse)? Bijection because in the associated commutative diagram we get mappings between sets which should be perhaps bijections?
 A: If $F,G:\mathbf{C}\to\mathbf{D}$ are functors, a natural transformation $\alpha:F\to G$ has an inverse if and only if the component $\alpha_A$ is an isomorphism for each $A\in\mathbf{C}$. In $\mathbf{Set}$ the isomorphisms are precisely the bijections. So "natural bijection" just means a natural isomorphism in the case where the target category is $\mathbf{Set}$.
In particular homsets are sets, so natural isomorphisms involving them may be called natural bijections.
A: It is, as you said, a bijective natural transformation and the naturality condition is one and the same as the defining condition for it to be a natural transformation.
In the following, I will use the conventions that for a category $$:

*

*$||$ = the objects of $$,

*for $X, X' ∈ ||$, $(X,X')$ = the set of arrows/morphisms $X → X'$,

along with the definitions

* 

*$^+$ is the dual category:

*

*$|^+| = ||$,

*for $X, X' ∈ ||$, $^+(X,X') = (X',X)$,

*identities in $^+$ are the same as those in $$,

*compositions in $^+$ are the reverse of those in $$.



*$×$ is the direct product category:

*

*$|×| = ||×||$,

*for $(X,Y),(X',Y') ∈ |×|$, $(×)((X,Y),(X',Y')) = (X,X')×(Y,Y')$.



For there to be an adjunction relation between categories $$ and $$ with functors
$$:  → , :  → $$
means that you have polymorphic operators
$$⋀: (X, Y) → (X, Y),$$
$$⋁: (X, Y) → (X, Y),$$
such that for ($X ∈ ||$, $Y ∈ ||$)
$$f: X → Y ⇒ ⋁⋀f = f,$$
$$g: X → Y ⇒ ⋀⋁g = g,$$
thereby making them inverses, such that the following naturality conditions hold
$$⋀(l∘f∘k) = l∘⋀f∘k, ⋁(l∘g∘k) = l∘⋁g∘k,$$
where
$$k: X' → X, l:Y → Y'.$$
Make the parametrization of the polymorphic operators explicit $⋀_{X,Y}$ and $⋁_{X,Y}$ and there's your natural transformation.
The natural transformations are thus $⋀:  → $, $⋁:  → $, where the functors
$$,: ^+× → $$
are given by
$$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): f ∈ (X,Y) ↦ (l∘f∘k) ∈ (X',Y')),$$
and
$$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): g ∈ (X,Y) ↦ (l∘g∘k) ∈ (X',Y')).$$
The defining conditions for $⋀$ and $⋁$ to be natural transforms are:
$$⋀_{X',Y'}∘(k,l) = (k,l)∘⋀_{X,Y},$$
$$⋁_{X',Y'}∘(k,l) = (k,l)∘⋁_{X,Y},$$
where
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) = (X',X) × (Y,Y'),$$
i.e. $k: X' → X$ and $l: Y → Y'$.
When applied to
$$f ∈ (X,Y), g ∈ (X, Y),$$
i.e. to $f: X → Y$ in $$ and $g: X → Y$ in $$, this yields the result:
$$⋀_{X',Y'} (k,l)(f) = (k,l)\left(⋀_{X,Y} f\right),$$
$$⋁_{X',Y'} (k,l)(g) = (k,l)\left(⋁_{X,Y} g\right),$$
or, upon application of the definitions:
$$⋀(l∘f∘k) = l∘⋀f∘k, ⋁(l∘g∘k) = l∘⋁g∘k.$$
