Let $F : C \to D, \ \ G : D \to C$ be two functors we're trying to prove are quasi-inverse to each other.
Suppose I've proved that $F \circ G(x) \simeq \text{id}_D(x) = x$ via the collection of isomorphisms $\alpha_x$. In my particular problem $\alpha_x$ happens to be the unique such isomorphism.
How do I then show naturality or $\alpha_y \circ F\circ G(f) = f \circ \alpha_x$ for every $f : x \to y$ in $C$?
This is a question I had after another question and partial answer, if that provides any necessary background info.
The statement : "if $F:C\to C$ is a functor and for every $x$ there is a unique isomorphism $\alpha_x : F(x)\to x$; then $(\alpha_x)$ is a natural isomorphism" is not true, so unless you have some more hypotheses we can't help you.
Here's a counterexample : let $C$ be a category with a unique object $*$ and two morphisms : $x,id$ with $x\circ x = x$. Then there is a unique isomorphism $*\to *$.
Now let $F$ be the functor that sends $x$ to $id$ (the rest is forced by the definition of functor- you may check that it does define a functor). Clearly it satisfies the hypotheses of the claim (because there is a unique isomorphism $(F(*)=*)\to *$), but this isomorphism is not natural, because $id\circ F(x) = id\circ id = id$ while $x\circ id = x \neq id$