The identity natural transformation $\operatorname {\mathrm{id}}_{F}\colon F\Rightarrow F$ of $F$ is the natural transformation consisting of the collection
\[ \left\{ (\operatorname {\mathrm{id}}_{F})_{A}\colon F(A)\to F(A)\right\} _{A\in \operatorname {\mathrm{Obj}}(\mathcal{C})} \]
defined by
\[ (\operatorname {\mathrm{id}}_{F})_{A}\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{id}}_{F(A)} \]
for each $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$.
The naturality condition for $\operatorname {\mathrm{id}}_{F}$ is the requirement that, for each morphism $f\colon A\to B$ of $\mathcal{C}$, the diagram
commutes. This follows from unitality of the composition of $\mathcal{D}$, as we have
\begin{align*} F(f)\circ \operatorname {\mathrm{id}}_{F(A)} & = F(f)\\ & = \operatorname {\mathrm{id}}_{F(B)}\circ F(f),\\ \end{align*}
where we have applied unitality twice.
Let $A$ and $B$ be monoids and let $f,g\colon A\rightrightarrows B$ be morphisms of monoids. Applying the delooping construction of 
, we obtain functors $\mathsf{B}{f},\mathsf{B}{g}\colon \mathsf{B}{A}\rightrightarrows \mathsf{B}{B}$. We then have
\[ \operatorname {\mathrm{Nat}}(\mathsf{B}{f},\mathsf{B}{g})\cong \left\{ b\in B\ \middle |\ \begin{aligned} & \text{for each $a\in A$, we}\\ & \text{have $bf(a)=g(a)b$}\end{aligned} \right\} . \]
Unwinding the definitions in this case, we see that a transformation $\alpha $ from $\mathsf{B}{f}$ to $\mathsf{B}{g}$ consists of a collection
\[ \left\{ \alpha _{\bullet }\colon \bullet \to \bullet \right\} _{\bullet \in \operatorname {\mathrm{Obj}}(\mathsf{B}{A})} \]
of morphisms of $\mathsf{B}{B}$ indexed by $\operatorname {\mathrm{Obj}}(\mathsf{B}{A})$. Since $\operatorname {\mathrm{Obj}}(\mathsf{B}{A})=\mathrm{pt}$ and the morphisms of $\mathsf{B}{B}$ are precisely the elements of $B$, it follows that $\alpha $ corresponds precisely to the data of an element $b\in B$. Now, a transformation $[b]\colon \mathsf{B}{f}\Rightarrow \mathsf{B}{g}$ is natural precisely if, for each $a\in \operatorname {\mathrm{Hom}}_{\mathsf{B}{A}}(\bullet ,\bullet )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}A$, the diagram
commutes. Unwinding the definitions, we see that this diagram is given by and hence corresponds precisely to the condition $g(a)b=bf(a)$.
