Executive summary: the isomorphism is not particularly mysterious. It arises from the obvious choice of a trace from the group algebra of $G$ to the group algebra of $H$, given by the formula
$$\sum_{g \in G} c_g g \mapsto \sum_{h \in H} c_h h,$$ which induces an isomorphism from $RG$ onto $\mathrm{Hom}_{RH}(RG,RH)$ (here $R$ is an arbitrary commutative ring), and hence an isomorphism of the functors from induction (tensor product with $RG$) to co-induction (tensor product with $\mathrm{Hom}_{RH}(RG,RH)$).
Now for the details. First, it may help contextualize things to ask a somewhat more general question: given a homomorphism $R \to S$ of rings, when are the induction $S \otimes_R \bullet$ and co-indunction $\mathrm{Hom}_R(S,\bullet)$ functors isomorphic, and when this is the case, how can one describe the set of all such isomorphisms?
There are two relatively obvious conditions that $S$ must satisfy if this is the case. First, since $S \otimes_R \bullet$ is right-exact, if there is such an isomorphism then $\mathrm{Hom}_R(S,\bullet)$ must be right-exact as well, implying that $S$ is projective as a left $R$-module. Second, since $S \otimes_R \bullet$ commutes with arbitrary direct sums, so must $\mathrm{Hom}_R(S,\bullet)$, implying that $S$ is finitely generated as a left $R$-module.
Suppose now that $S$ is projective and finitely generated as a left $R$-module. Then, just as for vector spaces over a field, the map $\mathrm{Hom}_R(S,R) \otimes_R M \to \mathrm{Hom}_R(S,M)$ given by $\phi \otimes m \mapsto [s \mapsto \phi(s) m]$ is a functorial isomorphism: this is true if $S=R$, hence for finitely generated free modules, hence for their summands, the finitely generated projective $R$-modules.
Summing up: an isomorphism from induction to co-induction can exist only if $S$ is finitely generated and projective as an $R$-module, and in this case it is the same thing as an isomorphism of $(S,R)$-bimodules
$$S \longrightarrow \mathrm{Hom}_R(S,R).$$ In fact, there is an isomorphism
$$\mathrm{Hom}_{R,R}(S,R) \to \mathrm{Hom}_{S,R}(S, \mathrm{Hom}_R(S,R))$$ given by sending a "trace" $t: S \to R$ to the map $s \mapsto [s' \mapsto t(s' s)].$ So what we are looking for is a trace from $S$ to $R$ (that is, a morphism of $R,R$-bimodules) such that the corresponding map from $S$ to $\mathrm{Hom}_R(S,R)$ is an isomorphism (necessarily of $S,R$-bimodules).
Now we specialize all this to group algebras. Thus let $H \leq G$ be a subgroup of the finite group $G$, and let $R$ be a commutative ring (we could more generally work with a map from one group $H$ to another $G$, but there is a simple obstruction: we need $RG$ to be a projective $RH$-module, which fails e.g. if the map is trivial and the order of $H$ is not invertible in $R$). Thus $RG$ is a finitely generated free, and hence projective, $RH$ module. An obvious candidate for a trace from $RG$ to $RH$ is the projection map
$$\sum_{g \in G} c_g g \mapsto \sum_{h \in H} c_h h.$$ One checks that this induces an isomorphism of $RG$ onto $\mathrm{Hom}_{RH}(RG,RH)$, mapping $g \in G$ to the function $$g' \mapsto \begin{cases} g' g \quad \hbox{if $g' g \in H$, and} \\ 0 \quad \hbox{ else.} \end{cases}$$ and hence an isomorphism from induction onto co-induction. Given an $RH$-module $M$ this map may be written explicitly as follows: it is given on $g \otimes m \in RG \otimes M$ by
$$g \otimes m \mapsto \left[ g' \mapsto \begin{cases} g' g m \quad \hbox{if $g' g \in H$, and} \\ 0 \quad \hbox{else.} \end{cases} \right]$$ In the case in which the order of $H$ is invertible in $R$, the inverse of this map is recorded in the first answer you linked to in your question.
This answer is long enough as it is, but I would like to mention that although the unit-counit compositions you mention in your question are not necessary for understanding the induction-coinduction isomorphism, they are useful for analyzing relative projectivity of modules over group algebras, and in particular for the theory of vertices and sources and the Green correspondence, as indicated in the next paragraph.
The composition of the unit for the co-induction and the co-unit for the induction produces an endomorphism of the identity functor on $RG$-mod, or in other words a central element of $RG$. This central element can be computed explicitly: it is simply the index $[G:H]$. As a consequence, if this index is invertible in $R$ then every $RG$-module is a summand of its induction-restriction to $H$, and in particular is relatively $H$-projective. So in case $R$ is a field of characteristic $p$, every $RG$-module is relatively $H$-projective provided $H$ contains a Sylow $p$-subgroup of $G$.
We have $$\mathsf{Hom}(A,B)\stackrel{\eta_B\circ-}{\to}\mathsf{Hom}(A,GFB)\stackrel{\varphi_{A,FB}}{\cong}\mathsf{Hom}(FA,FB)$$ natural in $A$ and $B$ with $\eta$ being the unit. This would immediately make fullness and faithfulness of $F$ be equivalent to $\eta$ being an isomorphism (via Yoneda) as long as we can show that $Ff=\varphi(\eta\circ f)$. Specifically, we'd have $$\mathsf{Hom}(A,B)\stackrel{F}{\cong}\mathsf{Hom}(FA,FB)\stackrel{\varphi^{-1}_{A,FB}}{\cong}\mathsf{Hom}(A,GFB)$$ is equal to $$\mathsf{Hom}(A,B)\stackrel{\eta_B\circ-}{\to}\mathsf{Hom}(A,GFB)$$ meaning it's an isomorphism and Yoneda (like all fully faithful functors) reflects isomorphisms. The other direction is obvious: $F$ would be a composition of isomorphisms.
The necessary equality is immediate by naturality of $\varphi$ (in $A$ as above) which lets us reduce to $Fid=id=\varphi(\eta)$ and indeed $\eta$ is defined as (or provably equivalent to) as $\varphi^{-1}(id)$. (In the cases where $\eta$ is primitive, then the definition of $\varphi$ and its inverse will make the above equation fairly obvious.)
Best Answer
Expanding a little bit on Derek Elkins' comment above.$\newcommand{\D}{\mathcal D}$ $\newcommand{\C}{\mathcal C}$
Assume that $$\varphi \colon \D(F(X),A) \stackrel{\sim}{\longrightarrow}\C(X,G(A))$$ is the natural isomorphism of the adjunction.
The natural isomorphism you described, namely $$\D(A,B) \stackrel{G}{\longrightarrow}\C(G(A),G(B)) \stackrel{\varphi^{-1}}{\longrightarrow} D(F(G(A)),B)$$ associates to each $f \colon A \to B$ the morphism $\epsilon_B\circ F(G(f)) \colon F(G(A)) \to B$ (this comes from the fact that $\varphi^{-1}(g)= \epsilon \circ F(g)$).
Using naturality of $\epsilon$ it follows that $$\epsilon_B \circ F(G(f))=f\circ\epsilon_A$$ that means the natural isomorphism is given by precomposition $\epsilon_A$, that is it is represented by $\epsilon_A$ via the Yoneda embedding.
Since embeddings reflect isomorphism, and since the Yoneda embedding is an embedding, it follows that $\epsilon_A$ must be an isomorphism.