TITLE: What is known about relative adjunctions?
QUESTION [10 upvotes]: I couldn't add to the well-written $n$Lab page about relative adjoints, so let me start taking that definition for granted.
I have a few questions about how the classical results on adjoints remain true, even in this fairly asymmetric setting.
I would like to understand better the meaning of this structure (e.g. when, how and why does it arise "in practice"). I collect here a few questions I have no clue how to attack.

Let's assume that there are relative adjunctions $f\dashv_j g$ and $f' \,{}_{j'}\!\!\dashv g'$; we can then arrange a pair of commutative triangles
$$
\begin{array}{ccc}
\bullet & \overset{f}\to & \bullet\\
\hskip-4mm g\uparrow & \nearrow \\
\bullet && 
\end{array}
\qquad \epsilon : fg \Rightarrow j
$$
$$
\begin{array}{ccc}
&&\bullet \\
&\nearrow& \hskip5mm \uparrow g'\\
\bullet &\underset{f'}\to& \bullet
\end{array}
\qquad \eta : j' \Rightarrow g'f'
$$
assume, now, that there exists a natural transformation $j\Rightarrow j'$; under which assumptions the resulting pasting square is filled by an invertible 2-cell?
Assume that 
$$
g_1 \dashv_{j_1} f \;{}_{j_2}\!\!\dashv g_2
$$
are relative adjoints; in this particular case it is possible to build a 2-cell $j_2 g_1\Rightarrow g_2 j_1$ pasting the unit of $f \;{}_{j_2}\!\!\dashv g_2$ with the counit of $g_1 \dashv_{j_1} f$. Under which conditions is this 2-cell an iso? In the same situation, the composition $j_1 j_2$ is parallel to $f$; is there a "comparison 2-cell" to/from $f$, and if there is, under which conditions it is invertible?
The $n$Lab says that $f \; {}_j\!\!\dashv g$ if $f\cong \text{LIFT}_gj$ (absolute lift); is this an iff? This relation means that $g$ uniquely determines $f$; the converse is not true in general: does it mean that there can be two non-isomorphic $g,g'$ such that $f \; {}_j\!\!\dashv g$ and $f \; {}_j\!\!\dashv g'$? Is there an instructive example of this? What is the structure, if any, of the class $\{g\mid f \; {}_j\!\!\dashv g\}$?
If $j$ is an isomorphism and $f \;{}_{j}\!\!\dashv g$, then this means that $f\dashv j^{-1}g$; is this condition any different from the particular case $f\dashv g = f \;{}_1\!\!\dashv g$? Is there a nontrivial, instructive example of this situation? I am confused by the loss of uniqueness outlined in the previous point...
One of the reasons why relative adjunctions are useful is that they model weighted co/limits: writing the weight in a suitable form, one has that $j\otimes f \dashv_j \hom(f,1)$, which is, well, quite obvious if you write
$$
{\bf hom}(j\otimes f,1) \cong {\bf hom}(j, \hom(f,1))
$$
how far can this analogy be pushed? (write $j\otimes(j'\otimes f) \cong \dots$: what does it mean for the associated relative adjunction?)
What is the meaning of relations like $1_A \dashv_j g$ for suitable functors? such a situation gives an isomorphism of profunctors $\hom \cong \hom(j,g)$, so in some sense finding "all $j$-right adjoints of the identity" means finding all factorizations of the identity profunctor $\hom_A$. How far can this equivalence be pushed? What is the formal meaning of this analogy in terms of profunctor theory?
What does the invertibilty of the relative unit $\eta : j\Rightarrow gf$ imply for $g,f$ ($f$ is "relatively fully faithful"?!). Dual question for the counit.

REPLY [3 votes]: Using the answer to (7), if the natural transformation $J \Rightarrow J'$ is invertible, and $G$ and $F'$ are fully faithful, then the pasting square will commute up to natural isomorphism (assuming $J$ is dense and fully faithful).

Again using the answer to (7), if $F$ and $G_1$ are fully faithful, then the composite natural transformation will be invertible (assuming $J$ is dense and fully faithful). As far as I can tell, the second part of the question is ill-formed.

Yes, $F$ is a $J$-relative left-adjoint to $G$ if and only if $f \cong \mathrm{lift}_G J$ and this left lifting is absolute (this is an easy exercise). An example of non-unique relative right adjoints is given in Ulmer's Properties of Dense and Relative Adjoint Functors: let $J : \mathrm{AbGrp}_f \to \mathrm{AbGrp}$ be the inclusion of finite Abelian groups in all Abelian groups and let $L : \mathrm{AbGrp}_f \to \mathrm{Vect}_{\mathbb Q}$ be the zero functor. Then $L$ is $J$-relative left adjoint both to $0 : \mathrm{Vect}_{\mathbb Q} \to \mathrm{AbGrp}$ and to the forgetful functor $V : \mathrm{Vect}_{\mathbb Q} \to \mathrm{AbGrp}$. I'm not aware of any nice structure on the set of $J$-relative right adjoints to a given functor.

Adjunctions relative to isomorphisms are the same as adjunctions relative to the identity. There is no lack of uniqueness here, since isomorphisms are dense, and $J$-relative right adjoints are unique when $J$ is dense.

Weighted colimits as relative adjunctions are treated in §4 of Street–Walter's Yoneda structures on 2-categories, where they are called "weak indexed colimits". See there for various useful lemmas making use of this characterisation.

If the identity on $\mathbf A$ is $J$-relative left coadjoint to $G : \mathbf B \to \mathbf A$, then there is a natural isomorphism $\mathbf A(a, Jb) \cong \mathbf A(a, Gb)$ which, by Yoneda, just says that $J \cong G$.

Assume $J$ is dense and fully faithful. If $L$ is $J$-relative left adjoint to $R$, then the unit $\eta : J \Rightarrow RL$ is an isomorphism if and only if $L$ is fully faithful (easy exercise). Dually, assume $J$ is codense and fully faithful. If $L$ is $J$-relative left coadjoint to $R$, then the counit $\epsilon : LR \Rightarrow J$ is an isomorphism if and only if $R$ is fully faithful.