First attempt (disregard it and go to the second):
I'm trying to prove that $[\mathscr A,\mathscr B]^{op}\cong [\mathscr A^{op},\mathscr B^{op}]$.
I started with defining a functor $\mathscr F:[\mathscr A,\mathscr B]^{op}\to[\mathscr A^{op},\mathscr B^{op}]$ as follows.
On objects: $(F^{op}:\mathscr B\to \mathscr A)\mapsto (\mathscr F(F^{op}):\mathscr A^{op}\to\mathscr B^{op})$ where the functor $F(F^{op}):\mathscr A^{op}\to\mathscr B^{op}$ sends an object $A$ to $F(A)$ and a morphism $(f^{op}:A'\to A)$ to $(F^{op}(f^{op})=F(f)^{op}:F(A')\to F(A))$.
On morphisms: not sure. Given a natural transformation $\alpha=(\alpha_B:F^{op}(B)\to G^{op}(B))_{B\in\mathscr B}$ between $F$ and $G$, we need to define a natural transformation $\mathscr F(\alpha)=(\mathscr F(\alpha)_A:\mathscr F(F^{op})(A)\to \mathscr F(G^{op})(A))_{A\in\mathscr A^{op}}=(\mathscr F(\alpha)_A:F(A)\to G(A))_{A\in\mathscr A}$. I don't quite see how to define this in terms of $\alpha_B$. Each $\alpha_B$ is an arrow in $\mathscr A$, but we need $\mathscr F(\alpha)_A$ to be an arrow in $\mathscr B$. I thought about taking $\mathscr F(\alpha)_A$ to be $F(\alpha_A)$, but $\alpha_A$ (for $A\in \mathscr A$) doesn't make sense.
Another attempt:
Define $\mathscr F:[\mathscr A,\mathscr B]^{op}\to[\mathscr A^{op},\mathscr B^{op}]$ as follows.
On objects: given a functor $F:\mathscr A\to \mathscr B$, assign to it the functor $F^{op}:\mathscr A^{op}\to\mathscr B^{op}$. (This $F^{op}$ is defined on objects by $F^{op}(A)=F(A)$ and on morphisms by $F^{op}(f^{op}:A'\to A)=F(f)^{op}$.)
Now take a morphism in $[\mathscr A,\mathscr B]^{op}$. It is a natural transformation $\alpha^{op}: G\to F$ (where $F,G:\mathscr A\to\mathscr B$ are functors). The naturality of $\alpha^{op}$ says this: for any arrow $g:A\to A'$ in $\mathscr A$, we have
$$\alpha_{A'}^{op}\circ G(g)=F(g)\circ \alpha_A^{op}.$$
We need to assign to $\alpha^{op}$ a natural transformation $\mathscr F(\alpha^{op}): G^{op}\to F^{op}$. So for each $A\in\mathscr A$ we need to define $F(\alpha^{op})_A=:\beta_A$. The following naturality condition must hold: for any arrow $f^{op}:A'\to A$ in $\mathscr A^{op}$, $$F(f)\circ_{op} \beta_{A'}=\beta_A\circ_{op} G(f)^{op}.$$
I was thinking about $\beta=(\alpha^{op})^{op}=\alpha$. But $\alpha_A$ is an arrow $F(A)\to G(A)$, whereas $\beta_A$ has to be an arrow $G(A)\to F(A)$. And $\beta_A=\alpha_A^{op}$ doesn't work either because if you substitute $\alpha$ for $\beta$ in the naturality condition (second display), it doesn't check out.
This isn't quite right. An object of $[\mathscr A,\mathscr B]^\mathrm{op}$ is just a functor $F:\mathscr{A}\to \mathscr{B}$, which we can call $F^\mathrm{op}$ to distinguish which category it lives in if we wish. It's the natural transformations that are reversed. In your definition $F(f)^\mathrm{op}$ doesn't parse, since $F$ is assumed to have domain $\mathscr B$.
So let's start again. Send a functor $F:\mathscr A\to \mathscr B$ to the functor $F^\mathrm{op}:\mathscr A^\mathrm{op}\to \mathscr B^\mathrm{op}$ sending $a$ to $F(a)$ and a morphism $f^\mathrm{op}\to a'$ in $\mathscr A^\mathrm{op}$ to $F(f)^\mathrm{op}:F(a')\to F(a)$ in $\mathscr B^\mathrm{op}$. Can we get somewhere sensible with the action on natural transformations now?