Let $\sigma : F \rightarrow F'$ be a natural transformation where $F, F' : X \rightarrow A$.
If $\phi : A(Fx, a) \cong X(x, Ga)$, and $(\sigma_x)^*=A(\sigma_x, a) :A(F'x,a) \rightarrow A(Fx, a)$ and $\phi'^{-1} : X(x, G'a) \cong A(F'x, a)$,
How do I apply the Yoneda lemma to the composite transformation: $u=\phi \circ (\sigma_x)^* \circ \phi'^{-1}$?
I'm trying to follow the proof in the image below to prove that the graph at the very bottom of the attached picture is commutative.
From Mac Lane's Category Theory:
In the included picture, $$(4)\text{ is } \langle F,G, \phi \rangle:X \rightarrow A, \langle F', G', \phi' \rangle: X \rightarrow A \text{ and } \sigma : F \rightarrow F', \tau: G' \rightarrow G$$
Fix $a$.
The diagram and the formula shows that there is a unique arrow $f_x :X(x,G'a) \to X(x,Ga)$ making the wanted diagram commute.
Now if you prove that this $f_x$ (namely $\varphi\circ (\sigma_x)^* \circ \varphi'^{-1}$; $\varphi$ and $\varphi'$ should actually be indexed by $x$ and $a$, and so for our purposes at least by $x$, since $a$ is fixed, but McLane often doesn't index these) is the component of a natural transformation $f:X(-, G'a) \to X(-, Ga)$, the Yoneda lemma tells you that this map is induced by a unique arrow $\tau_a: G'a\to Ga$.
This $\tau_a$, he claims, is then the component of a natural transformation $\tau : G'\to G$ (it's actually easy to see because if you follow the proof of the Yoneda Lemma you can actually make this $\tau_a$ explicit, and it is clearly natural in $a$, since $\varphi, \varphi'$ are).
I think this is what is meant by "apply the Yoneda Lemma to $\varphi\circ (\sigma_x)^* \circ \varphi'^{-1}$", though the way he writes it is a bit unclear as it should be more something like $\varphi\circ (\sigma_{-})^* \circ \varphi'^{-1}$ or ($x\mapsto \varphi\circ (\sigma_x)^* \circ \varphi'^{-1}$) (indeed, $\varphi\circ (\sigma_x)^* \circ \varphi'^{-1}$ is not by itself a natural transformation, it's only the component of such a transformation evaluated at $x$)