Uniqueness of adjoint functors up to isomorphism

The bijection $\hom(C,GD) \cong \hom(C,G'D)$ is natural in $C$, thus it is induced by an isomorphism $GD \cong G'D$ (Yoneda Lemma). But it is also natural in $D$, and since the Yoneda embedding is faithful, this means that $GD \cong G'D$ is natural in $D$. More details:

If $D \to E$ is a morphism, then

$\begin{array}{ccc} GD & \rightarrow & G'D \\ \downarrow & & \downarrow \\ GE & \rightarrow & G'E \end{array}$

commutes iff for every $C$ the diagram

$\begin{array}{ccc} \hom(C,GD) & \rightarrow & \hom(C,G'D) \\ \downarrow & & \downarrow \\ \hom(C,GE) & \rightarrow & \hom(C,G'E) \end{array}$

commutes.

This seems to be one of those cases where the proof from scratch is the easy, and you don't need to know (explictly) anything about Yoneda.

1). You know there is a natural isomorphism given by $\varphi_{A,D} :$hom$(A,GD)\rightarrow $hom$(A,G'D)$

2). Set $A=GD$, $\varphi _{GD,D}(1 _{GD})=\tau _{D}$. Then, by what follows, the $\tau _{D}$ are the components of the required natural isomorphism $G\overset{\cdot }{\rightarrow}G'$

3) Write

\begin{array}{ccc} \hom(GD,GD) & \rightarrow & \hom(GD,G'D) \\ \downarrow & & \downarrow \\ \hom(GD,GD_{1}) & \rightarrow & \hom(GD,G'D_{1} ) \end{array} Then if $f:D\rightarrow D_{1}$, follow $1_{GD}$ around the square. This gives $G'f\circ \tau _{D}=\varphi _{GD,D_{1}}(Gf)$.

4). Write \begin{array}{ccc} \hom(GD_{1},GD_{1}) & \rightarrow & \hom(GD_{1},G'D_{1}) \\ \downarrow & & \downarrow \\ \hom(GD,GD_{1}) & \rightarrow & \hom(GD,G'D_{1} ) \end{array}

and follow $1_{GD_{1}}$ around the square. This gives $\varphi _{GD,D_{1}}(Gf)= \tau _{D_{1}}\circ Gf$

5). Combining the previous two items, you get $G'f\circ \tau _{D}=\tau _{D_{1}}\circ Gf$, so $\tau $ is a natural transformation. Now, since $\varphi$ is an isomorphism, natural in $A$ and $D$, for each $D$, $\varphi _{GD,D}(1 _{GD})=\tau_{D}$ is an isomorphism, which implies $\tau $ itself is.

We actually have a composite of natural isomorphisms $\tau:\hom(\bullet,G\bullet)\to \hom(F\bullet,\bullet)\to \hom(\bullet,G'\bullet)$.

In more details, $\tau$ has components $\tau_{C,D}:\hom(C,GD)\to\hom(C,G'D)$ satisfying the naturality condition: for arrows $\gamma:C_1\to C$, $\ \delta:D\to D_1\,$ and $\, f:C\to GD$ we have $$\tau(G\delta\circ f\circ\gamma)=G'\delta\circ\tau(f)\circ\gamma\,. $$ Now applying the Yoneda lemma basically means to consider the case when $C=GD$ and apply $\tau$ to $1_{GD}$. It leads to a $GD\to G'D$ arrow, verify (using the above equation) that it's natural in $D$, and find its inverse (e.g. by applying the same to $C=G'D$ and $\tau^{-1}$.