This question is related to point $(1)$ in this answer:
https://math.stackexchange.com/a/190402/229776
Let $(\mathsf{C}, \otimes, 1, \alpha, l ,r, s)$ be a symmetric monoidal category, $(\mathsf{C}_s, \otimes_s, 1_s)$ a strict monoidal category and $F\colon\mathsf{C}\to\mathsf{C}_s$ a (strong) monoidal equivalence together with $(J_{X,Y}\colon F(X)\otimes_s F(Y)\to F(X\otimes Y))_{X,Y \in \mathsf{C}}$
The aforementioned answer claims that $\mathsf{C}_s$ is also a symmetric monoidal category by symmetry induced by $s$. For each $A \in \mathsf{C}_s$, choose $X_A \in \mathsf{C}$ and an isomorphism $f_A\colon F(X_A)\to A$. It seems a canonical choice for a symmetry isomorphism in $\mathsf{C}_s$ is $(f_B\otimes_s f_A)\circ J^{-1}_{X_B,X_A}\circ F(s_{X_A,X_B})\circ J_{X_A,X_B} \circ (f^{-1}_A\otimes_s f^{-1}_B)$, which we denote by $t_{A,B}$. It's not hard to prove that $t$ is natural and one of the axioms for symmetry isomorphism (relating unit and symmetry) is straightforward due to $\mathsf{C}_s$ being strict. However, proving that the diagram
commutes feels beyond me.
Of course, by symmetry in $\mathsf{C}$ and functoriality, the diagram
commutes, so one needs to use this.
My idea is to use the commutative hexagon in the middle and to build commutative squares around its edges using paths of the form: $$J_{X_A\otimes X_B,X_C}\circ (J_{X_A,X_B}\otimes 1_{F(X_C)})\circ((f^{-1}_A\otimes f^{-1}_B)\otimes f^{-1}_C)\colon (A\otimes B)\otimes C\to F((X_A\otimes X_B)\otimes X_C).$$ It feels like I'm in the right direction, but I still can't make it work. If I have a say, I would like an answer to help me with the details, since it's what I'm having a trouble with.