Does the following commutative diagram follow from the fact that a braiding is a natural transformation? A braided monoidal category is a monoidal category with a braiding (a commutativity constraint $\gamma_{A,B}: A \otimes B \to B \otimes A$) which satisfies the hexagon identities:
$$
\gamma_{A, B \otimes C} = (1 \otimes \gamma_{A,C})(\gamma_{A,B} \otimes 1) \quad (1) \\
\gamma_{A \otimes B, C} = (\gamma_{A,C} \otimes 1) (1 \otimes \gamma_{B,C}). \quad (2)
$$
In a monoidal category, we have two functors $F,G: \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, $F(A, B) = A \otimes B$,  $G(A,B) = B \otimes A$. The braiding $\gamma_{A,B}: A \otimes B \to B \otimes A$ is a natural transformation: $\gamma: F \to G$.
Does the following commutative diagram follow from the fact that a braiding is a natural transformation?
$$\begin{array}
{cccc} A \otimes B \otimes C  & \stackrel{\gamma_{A, B \otimes C}}{\longrightarrow} & B \otimes C \otimes A \\
\downarrow{\gamma_{B,C}} & & \downarrow{\gamma_{B,C}} \\
A \otimes C \otimes B & \stackrel{\gamma_{A, C \otimes B}}{\longrightarrow} & C \otimes B \otimes A  
\end{array}
$$
I tried to use the definition of natural transformation. Which morphism $f: \mathcal{C} \times \mathcal{C} \to \mathcal{C} \times \mathcal{C}$ should we take? If we take $f=(1, \gamma_{B,C}): (A, B \otimes C) \to (A, C \otimes B)$, then we have $F(f): A \otimes B \otimes C \to B \otimes C \otimes A$ given by $F(f)=1 \otimes \gamma_{B \otimes C}$, where $F$ is defined above. Similarly, we have $G(f): B \otimes C \otimes A \to C \otimes B \otimes A$ given by $G(f)= \gamma_{B \otimes C} \otimes 1$, where $G$ is defined above. Therefore we have the above commutative diagram. Is this correct?
Thank you very much.

Yes, that is because in your diagram the vertical arrows are in fact from left to right $A\otimes\gamma_{B,C}$ and $\gamma_{B,C}\otimes A$ and that the horizontal lines are the corresponding natural transformations, $\gamma_{A,-} : A\otimes- \to -\otimes A$.
More precisely, let $F:[M\mapsto A \otimes M, f\mapsto id_A \otimes f]$ and  $G:[M \mapsto M\otimes A,f\mapsto f \otimes id_A]$ then $\gamma_{A,-}$ is by definition a natural transformation from $F$ to $G$ and the diagram commutes indeed. 
