From the question Adjoint Squares:
Let $(F, G, η, ε)$ and $(F', G', η', ε')$ be adjunctions, $F: \mathscr C → \mathscr D$, and $F' : \mathscr C' → \mathscr D'$. There is a powerful isomorphism of functors $\mathrm{Nat}(F'-, -F)$ and $$\mathrm{Nat}(-G, G'-) : [\mathscr C, \mathscr C']^{\mathrm{op}} × [\mathscr D, \mathscr D'] → \mathrm{Set},$$ and in particular, for every $K : \mathscr C → \mathscr C'$ and $L : \mathscr D → \mathscr D'$ a bijection $$\mathrm{Nat}(F'K, LF) ≅ \mathrm{Nat}(KG, G'L)$$ known as the mate correspondence. For $α : F'K ⇒ LF$, we get the mate $β : KG ⇒ G'L$ as $G'Lε ◦ G'αG ◦ η'KG$, and the inverse is defined similarly: $β : KG ⇒ G'L$ gets sent to $ε'LF ∘ F'βF ∘ F'Kη$.
Say you had a natural isomorphism $\alpha\colon F'K\Rightarrow LF$ with inverse $\alpha^{-1}\colon LF\Rightarrow F'K$. The mate correspondence above gives a mate $\beta\colon KG\Rightarrow G'L$ for $\alpha$, and likewise $\alpha^{-1}$ has a mate $\gamma\colon G'L\Rightarrow KG$.
Is it the case that $\beta$ is an isomorphism, and $\beta^{-1}=\gamma$? I can write down $\beta$ and $\gamma$ in terms of $\alpha$ and $\alpha^{-1}$ and the appropriate units and counits based on the formulas quoted above, but trying to compose the resulting functors doesn't seem to lead anywhere since a counit precedes a unit, so the usual triangle identites for the unit-counit pairs does not apply.
As Arnaud D. points out in the comments, the question doesn't really work syntactically, since mate correspondence cannot be applied to inverses. You could still ask if it preserves invertibility, but the answer is again no.
To see this, notice that the adjunction isomorphism itself is a special case of the mate correspondence. Take $\mathrm{Id} ⊣ \mathrm{Id}$ on the terminal category for $F ⊣ G$ above, and write $F ⊣ G$ instead of $F' ⊣ G'$ for simplicity. That gives you $\mathrm{Nat}(FK, L) ≅ \mathrm{Nat}(K, GL)$ for all $K$, $L$ with domain $1$ ie. basically $\mathrm{Hom}(FK, L) ≅ \mathrm{Hom}(K, GL)$ for all objects $K$, $L$. But of course, the adjunction isomorphism doesn't usually translate isomorphisms to isomorphisms.
The good news is that taking mates does preserve composition in an appropriate sense. Call the triple $(α : F'K ⇒ LF, K, L)$ a ("left") morphism from $F ⊣ G$ to $F' ⊣ G'$. Then the mate correspondence just says that these are in bijection with what you'd get with the other reasonable definition (ie. "right" morphism, using $β : KG ⇒ G'L$). These morphisms can be composed in the obvious way, by pasting the corresponding squares, ie. setting $α'α := L'α ∘ α'K$, and you do have that $\overline{α'α} = \overline{α}\overline{α'}$, so that the categories of left and right morphisms are antiisomorphic. (If you want to prove this, you'll need to use naturality to bring $η'$ and $ε'$ next to each other so that you can cancel them out, but I'd recommend switching to string diagrams, with which it's immediate.)
Now if you take $K = L = \mathrm{Id}$, so that you're looking at adjunctions between a single pair of categores, you have that morphisms $F' ⇒ F$ correspond to morphisms $G ⇒ G'$ in a way that respects the ordinary composition of natural transformations, so in this special (but important) case, the inverses of natural transformations are preserved.