Does there exist a category $\mathcal{C}$ and contravariant functors $F:\mathcal{C}\to\mathcal{C}^{op}, G:\mathcal{C}^{op}\to\mathcal{C}$ such that $F$ is left ajoint to $G$ and $F\neq G^{op}$? In the case of $\mathcal{C}=Set$, I proved that there exist no such functors, because if exist $G$ preserve limits, and so is representable by adjoint functor theorem, and that $G=Hom(,A)$ has a left ajoint $F=Hom(,A)$ itself. But I can't come up with any answers to other cases. Partial solutions are welcomed. For example, when $\mathcal{C}$ is a topos, or $G$ is monadic, etc.
Here's a very simple example. Let $\mathcal{C}$ be any set with more than two elements, considered as a discrete category. Then $\mathcal{C}=\mathcal{C}^{op}$, and any permutation $F:\mathcal{C}\to\mathcal{C}^{op}=\mathcal{C}$ has an adjoint $G$ (on both sides!) given by the inverse. Any permutation that is not its own inverse then gives an example of an adjunction with $F\neq G^{op}$. 


Note that your analysis for the case $Set$ uses the fact that the cartesian product is symmetric monoidal. This suggests that you consider an example of a monoidal biclosed category which is nonsymmetric. (For example, consider presheaves on a small nonsymmetric monoidal category, equipped with the induced Day convolution.) To give a concrete instance, consider the poset of binary relations on a set $X$, where the monoidal product is relational composition. (Think of such relations as truthvalued functions $R(, ): X \times X \to \mathbf{2}$.) This product has an adjoint on each side, which we may denote $\Rightarrow$ and $\Leftarrow$: $$(R \Rightarrow S)(x, y) = \forall_z R(z, x) \to S(z, y)$$ $$(S \Leftarrow R)(x, y) = \forall_z R(y, z) \to S(x, z)$$ Then we have $ \Rightarrow S$ is adjoint to $S \Leftarrow $, since $$T \leq R \Rightarrow S \;\; \text{iff} \;\; T \circ R \leq S \;\;\text{iff}\;\; R \leq S \Leftarrow T$$ (modulo the fact that I may have switched the usual order of relational composition). But clearly these adjoint functors are nonisomorphic. 

