As you maybe remember, Isbell duality is an adjunction 
$$\mathcal O : [A°,Set] \leftrightarrows [A,Set]° : {\cal S}pec$$ 
as defined [here](https://ncatlab.org/nlab/show/Isbell+duality); since every functor $f : A\to B$ defines both 


1. a functor $f^\* : B\to [A°,Set]$ by $$(a,b)\mapsto B(fa,b)$$ and
2. a functor $f\_\* : B\to [A,Set]°$ by $$(a,b)\mapsto B(b,fa)$$


then it is reasonable to ask 



> 
> is it true that $\mathcal O \circ f^\*\cong f\_\*$ and ${\cal S}pec \circ f\_\* \cong f^\*$? 
> 
> 
> 


In other words the presheaf $B(f\\_\,,b)$ is exchanged with $B(b, f\\_\,)$ by $\mathcal O$, and similarly $f\_\*b$ becomes $f^\*b$ when it is post-composed with the ${\cal S}pec$ functor. Both $\cal O$ and ${\cal S}pec$ admit very explicit descriptions as
$$
\begin{gather\*}
{\cal O}(P)(A) = Nat(P, \hom(\\_\,,A))\\
{\cal S}pec(Q)(A) = Nat(Q, \hom(A,\\_\,)),
\end{gather\*}
$$
and while it seems to me that $\cal O$ has this property, I see no way to prove that
$$
Nat(B(b,f\\_\,), \hom(a,\\_\,)) \cong B(fa,b).
$$



Functors in Isbell duality exchange $f^*a$ and $f_*a$
Let's give names to these two separate conditions:
\begin{align}
Nat(B(b,f-),hom(a,-)) &\cong B(fa,b) \tag{1} \\
Nat(B(f-,b),hom(-,a)) &\cong B(b,fa) \tag{2} 
\end{align}
Neither of these must be true in general, though they hold under some natural conditions. I will treat (2) first, since (1) is completely dual.


As a counterexample, take $A = 1$ the terminal category, then $f$ corresponds to an object of $B$, and (2) reduces to a natural isomorphism
$$
Set(B(f,b),1) \cong 1 \cong B(b,f)
$$
which is false unless $f$ is a terminal object in $B$.


A sufficient condition for (2) is that $f : A \to B$ is both fully faithful and [dense](https://ncatlab.org/nlab/show/dense+functor), meaning that there are natural isomorphisms
\begin{align}
A(a,a') &\cong B(fa,fa') \tag{$f$ fully faithful}\\
B(b,b') &\cong Nat(B(f-,b),B(f-,b')) \tag{$f$ dense}
\end{align}
We then derive (2) in two steps:
$$
Nat(B(f-,b),hom(-,a)) \cong Nat(B(f-,b),B(f-,fa)) \cong B(b,fa)
$$
(As an aside, it is perhaps worth mentioning that fully faithful and dense functors play a role in the theory of [monads with arities](https://arxiv.org/abs/1101.3064).)


Essentially the same counterexample works with (1) (now taking $f : 1 \to B$ to be any non-initial object of $B$), and for a sufficient condition we can replace density with the [codensity](https://ncatlab.org/nlab/show/codense+functor)-like assumption that the functor $f\_\* : B \to [A,Set]^\circ$ is fully faithful, i.e., that there is an isomorphism $B(b,b') \cong Nat(B(b',f-),B(b,f-))$.


**Update (14 June)**: Restoring original answer (with expanded commentary) after realizing the flaw in the OP's counterargument. The isomorphisms ${\mathcal O} \circ f^\* \cong f\_\*$ and $Spec \circ f\_\* \cong f^\*$ cannot follow by preservation of (left/right) Kan extensions along (left/right) adjoints, since $f^\*$ is a *left* Kan extension while $f\_\*$ is a *right* Kan extension.


