Your notation is rather nonstandard; in my experience with these situations, $\eta$ is used for the unit of

Suppose $F \dashv G$ F^{op}$ with components unit $\eta_M: M \eta: 1_C \to GFM$ (which you have) F^{op} F$ and $\varepsilon$ for the counit with components $\varepsilon: F G N F^{op} \to N$ (which you don't have; yours is in the opposite direction)1_{C^{op}}$. I'm going to stick with the standard notation, so that the question becomes whether, in our situation, we have Since $\eta^{-1} = \varepsilon^{op}$.

Given \varepsilon$ is the unique transformation $F: C^{op\theta: F F^{op} \to C$ such that $F \dashv F^{op}$ and 1$ such that both unit and counit are invertible, we have $\eta^{-1} = \varepsilon^{op}$ if and only if

$$\eta F^{op} $1_{F^{op}} = (F^{op} \stackrel{\eta F^{op}}{\to} F^{op} F F^{op} \stackrel{F^{op} \varepsilon}{\to} F^{op} \stackrel{F^{op}\eta^{op}}{\to} F^{op} F F^{op}) = F^{op} \eta^{op}$$

where in the second equation we used a triangular identity. So stackrel{F^{op}\theta}{\to} F^{op})$$

the question becomes is whether we have $\eta F^{op} F^{op}\eta^{op} = F^{op} \eta^{op}$.

As I say, no(\eta F^{op})^{-1}$. Consider a

Take $C$ to be an commutative group $G$, considered as a one-object category $G$; choose with one object $G$ so that it contains a central element \bullet$. Here we may simply identify $g$ such that G^{op}$ with $g \neq g^{-1}$ (e.g., G$, i.e., the identity $G = 1_G \mathbb{Z}$ would do). Take $F: colon G \to G^{op}$ to G$ may be the functor that sends seen as a morphism $h$ to $h^{-1}$. The contravariant functor $F^{op}$ also takes because we have $h$ to 1_G(g h) = 1_G(h)1_A(g)$ by commutativity. Now take $h^{-1}$, F$ and the composite therefore $F^{op} F$ is F^{op}$ to be the identity . Denote the object of $G$ by on $\bullet$. We can use the central G$. Any element $g: \bullet \to u \bullet$ in G$ as a unit morphism $1_G \bullet \to F^{op} F$ because $g: \bullet \to F^{op}F = F^{op} F \bullet$ is natural (since $g$ is central), and $g$ can serve as a universal arrow the unit transformation (naturality also follows from commutativity). But then, as soon as $1_G \u$ is not equal to F^{op}F$ for trivial reasons. Then $$\eta F^{op} \bullet = \eta \bullet = g$$

and u^{-1}$, we have $$F^{op} \eta^{op} \bullet F^{op}u^{op} = F^{opu \neq u^{-1} g = g^{-1}$$ (u F^{op})^{-1}$. Therefore taking $G$ to be the additive group $\mathbb{Z}$ and therefore $\eta F^{op} u = 1 \neq F^{op} in \eta^{op}$. mathbb{Z}$, we reach a counterexample.

It's a good question; the answer is no.

Your notation is rather nonstandard; in my experience with these situations, $\eta$ is used for the unit of $F \dashv G$ with components $\eta_M: M \to GFM$ (which you have) and $\varepsilon$ for the counit with components $F G N \to N$ (which you don't have; yours is in the opposite direction). I'm going to stick with the standard notation, so that the question becomes whether, in our situation, we have $\eta^{-1} = \varepsilon^{op}$.

Given $F: C^{op} \to C$ such that $F \dashv F^{op}$ and such that both unit and counit are invertible, we have $\eta^{-1} = \varepsilon^{op}$ if and only if

$$\eta F^{op} = (F^{op} \stackrel{\eta F^{op}}{\to} F^{op} F F^{op} \stackrel{F^{op} \varepsilon}{\to} F^{op} \stackrel{F^{op}\eta^{op}}{\to} F^{op} F F^{op}) = F^{op} \eta^{op}$$

where in the second equation we used a triangular identity. So the question becomes whether we have $\eta F^{op} = F^{op} \eta^{op}$.

As I say, no. Consider a group as a one-object category $G$; choose $G$ so that it contains a central element $g$ such that $g \neq g^{-1}$ (e.g., $G = \mathbb{Z}$ would do). Take $F: G \to G^{op}$ to be the functor that sends a morphism $h$ to $h^{-1}$. The functor $F^{op}$ also takes $h$ to $h^{-1}$, and the composite $F^{op} F$ is the identity. Denote the object of $G$ by $\bullet$. We can use the central element $g: \bullet \to \bullet$ as a unit $1_G \to F^{op} F$ because $g: \bullet \to F^{op}F \bullet$ is natural (since $g$ is central), and $g$ can serve as a universal arrow $1_G \to F^{op}F$ for trivial reasons. Then

$$\eta F^{op} \bullet = \eta \bullet = g$$

and

$$F^{op} \eta^{op} \bullet = F^{op} g = g^{-1}$$

and therefore $\eta F^{op} \neq F^{op} \eta^{op}$.