Suppose that $C,D$ are symmetric monoidal categories and $F:C\rightarrow D$ is a nonsymmetric monoidal functor. 


1. Does it imply that there exists isomorphic functor $G$ which is a symmetric monoidal functor?
2. If not, then maybe additional assumption that $F$ is an equivalence should be imposed?


In other words what nice properties the 2-functor of inclusion $\mathbf{Symm}(Cat)\rightarrow \mathbf{Mon}(Cat)$ have? 


$\mathbf{Symm}(Cat)$:= 2-category of symmetric monoidal categories and symmetric monoidal functors.


$\mathbf{Mon}(Cat)$:= 2-category of monoidal categories and monoidal functors.



symmetry in monoidal categories
The answer to both questions is no, since a monoidal category can have several non-equivalent symmetric structures. For instance, the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces can be given the "usual" symmetry, 
$$c(x \otimes y) = y \otimes x,$$
or the "super" symmetry,
$$c(x \otimes y) = (-1)^{\lvert x \rvert \lvert y \rvert} y \otimes x.$$
Thus, the identity functor between the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces with the "usual" symmetry and the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces with the "super" symmetry is a monoidal equivalence, but not symmetric monoidal.


Edit:


With regard to classifying symmetric structures on monoidal categories, I don't think anything useful can be said in complete generality, but there are some interesting results related to the theory of Tannaka duality. Here's one such statement, which is a special case of a result of Deligne:



> 
> *Theorem.* Let $\mathcal{C}$ be a unitary symmetric fusion category over $\mathbb{C}$. Suppose further that the associated twist $\Theta\_X$ is equal to $\operatorname{Id}\_X$ for all objects $X \in \mathcal{C}$. Then $\mathcal{C}$ admits a unique fiber functor which identifies $\mathcal{C}$ with the category $\operatorname{Rep} G$ for some finite group $G$.
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Thus, given a unitary fusion category $\mathcal{C}$, the set of all unitary symmetric structures with trivial twist is the same as the set of groups $G$ for which $\mathcal{C} \cong \operatorname{Rep} G$ as unitary fusion categories. For a fixed $G\_0$, the collection of all $G$ with $\operatorname{Rep} G \cong \operatorname{Rep} G\_0$ in this way was classified by Etingof and Gelaki in their paper *[Isocategorical groups](http://arxiv.org/abs/math/0007196)*.


There is a generalization, originally due to Doplicher and Roberts, of Deligne's result where we remove the restriction on the twist. Let me describe the analogous result in the setting of unitary fusion categories.


We denote by $\operatorname{SVec}$ the category of finite dimensional super vector spaces, that is, the category I described in my original answer with the "super" symmetry. A finite supergroup is a group $G$ together with an element $x \in Z(G)$ with $x^2 = e$. A representation of a finite supergroup is a representation of the group $G$ on a super vector space $V$ such that $x$ acts by the identity on $V\_{\operatorname{even}}$ and by $-\operatorname{Id}$ on $V\_{\operatorname{odd}}$. Then we have the following result.



> 
> *Theorem.* Let $\mathcal{C}$ be a unitary symmetric fusion category over $\mathbb{C}$. Then $\mathcal{C}$ admits a unique fiber functor which identifies $\mathcal{C}$ with the category $\operatorname{Rep} (G, x)$ for some finite supergroup $(G, x)$.
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I don't know if anybody has extended the Etingof-Gelaki result to supergroups; I don't imagine this would be difficult.


I would welcome anybody who knows more about this than I do to expound further on the Doplicher-Roberts theorem and friends, or especially to correct anything I wrote here that is wrong!


