Part of proof of Yoneda's Lemma from Vakil

So, by part (a) you know that there exists a unique morphism $g\colon A\to A'$, such that for all objects $C$ the map $i_C\colon \operatorname{Mor}(C,A) \to \operatorname{Mor}(C,A')$ is given by $u\mapsto g\circ u$.
The inverse maps $i_C^{-1}\colon \operatorname{Mor}(C,A') \to \operatorname{Mor}(C,A)$ again satisfy the hypotheses of (a), so there exists a unique morphism $\tilde g\colon A'\to A$ such that for all objects $C$ the map $i_C^{-1}$ is given by $v\mapsto \tilde g\circ v$.

This is all you need from (a) to conclude: Putting $C=A$ we know that the composition $$ \operatorname{Mor}(A,A) \xrightarrow{i_A} \operatorname{Mor}(A,A')\xrightarrow{i_A^{-1}} \operatorname{Mor}(A,A) $$ is the identity. Hence, $$ \operatorname{id}_A = i_A^{-1}\bigl(i_A(\operatorname{id}_A)\bigr) = i_A^{-1}(g\circ \operatorname{id}_A) = \tilde g\circ (g\circ \operatorname{id}_A) = \tilde g\circ g. $$ Similarly, putting $C=A'$, the composition $$ \operatorname{Mor}(A',A') \xrightarrow{i^{-1}_{A'}} \operatorname{Mor}(A',A) \xrightarrow{i_{A'}} \operatorname{Mor}(A',A') $$ is the identity, so that $$ \operatorname{id}_{A'} = i_{A'}\bigl(i_{A'}^{-1}(\operatorname{id}_{A'})\bigr) = i_{A'}(\tilde g\circ \operatorname{id}_{A'}) = g\circ (\tilde g\circ \operatorname{id}_{A'}) = g\circ \tilde g. $$ But $\tilde g\circ g = \operatorname{id}_A$ and $g\circ \tilde g = \operatorname{id}_{A'}$ means that $g$ is an isomorphism with inverse $\tilde g$.