Is the functor $\mathrm{op} \colon \mathcal{C} \longrightarrow \mathcal{C}^\mathrm{op}$ equal to the identity? In Kashiwara-Schapira (Categories and abelian sheaves), it is written

*

*A contravariant functor from $\mathcal C$ to $\mathcal{C}'$ is a functor from $\mathcal{C}^\mathrm{op}$ to $\mathcal{C}'$.


*It is convenient to introduce the contravariant functor $\mathrm{op} \colon \mathcal{C} \longrightarrow \mathcal{C}^\mathrm{op}$
defined by the identity of $\mathcal{C}$.


*There is an isomorphism of categories
$$
  \mathrm{Fct}(\mathcal{C}, \mathcal{C}')^\mathrm{op} \simeq \mathrm{Fct}(\mathcal{C}^\mathrm{op}, \mathcal{C}'^{\mathrm{op}}),
  \quad F \longmapsto \mathrm{op} \circ F \circ \mathrm{op}.
$$
By 1), $\mathrm{op}$ is a functor from $\mathcal{C}^\mathrm{op}$ to $\mathcal{C}^\mathrm{op}$.
So, how could it be identity of $\mathcal{C}$? It perhaps the identity of $\mathcal{C}^\mathrm{op}$.
But in this case, in 3), we have $\mathrm{op} \circ F \circ \mathrm{op} = F$
which is not an element of $\mathrm{Fct}(\mathcal{C}^\mathrm{op}, \mathcal{C}'^{\mathrm{op}})$.
 A: For any category $\mathbf{C}$, there is the functor $\text{op}_\mathbf{C}\colon\mathbf{C}\to\mathbf{C}^\text{op}$ sending each object to itself (as the objects of a category and its opposite category are the same!), and sending any morphism $f\colon C\to D$ to the corresponding unique $f^\text{op}\colon D\to C$. While it is identity on objects, it is not that on morphisms (unless you only have identity morphisms in your category). We can then consider the functor $\text{op}_{\mathbf{C}^\text{op}}\colon\mathbf{C}^\text{op}\to\mathbf{C}$, and compose this with our previous functor and these satisfy $\text{op}_{\mathbf{C}^\text{op}}\circ\text{op}_\mathbf{C}=1_\mathbf{C}$ and $\text{op}_\mathbf{C}\circ\text{op}_{\mathbf{C}^\text{op}}=1_{\mathbf{C}^\text{op}}$. If there is even one non-identity morphism in $\mathbf{C}$, then $\text{op}_\mathbf{C}$ is not the identity functor.
How the claim in Kashiwara-Schapira should be read is $F\mapsto \text{op}_{\mathbf{C}'}\circ F\circ\text{op}_{\mathbf{C}^\text{op}}$, as this is a composite $$\mathbf{C}^\text{op}\xrightarrow{\text{op}_{\mathbf{C}^\text{op}}}\mathbf{C}\xrightarrow{F}\mathbf{C}'\xrightarrow{\text{op}_{\mathbf{C}'}}\mathbf{C}'^\text{op}.$$ What probably confused you is that they denote each of those opposite functors by $\text{op}$ without mentioning the category (as is usually the case). I'll let you figure out its inverse functor yourself to confirm this is indeed an isomorphism! :)
