Categorical definition of infinite symmetric product Let $(C,\otimes,I)$ be a symmetric monoidal category with coequalizers and directed colimits.
Fix some object $X$ and morphism $\tau\colon I\to X.$
Using $\tau$ one can construct a sequence of morphisms:
$$
I\rightarrow^{\tau}X\rightarrow^{\tau\circ\rho_X^{-1}} X\otimes X \rightarrow X^{\otimes 3} \ldots
$$
Let $(X,\tau)^{\infty}$ be a colimit of that diagram. 
Question: is it true(under which assumptions it is true?) that there exists a monomorphism $\eta\colon \Sigma_{\infty}\to Aut((X,\tau)^{\infty})$ induced by braidings on terms in the diagram? 
Here $\Sigma_{\infty}$ is a group of all permutations with finite support, we also assume that braiding $B_{X,X}$ is non-identity.
Why symmetric products?
Let us consider following diagram:
$$
\begin{array}{l}
I&\longrightarrow^{\tau}&X&\rightarrow^{\tau\circ\rho_X^{-1}} &X\otimes X& \longrightarrow& X^{\otimes 3}& \ldots\\
\downarrow_{id}&&\downarrow_{id}&&\downarrow_{Coeq(id,B_{X,X})}&&\downarrow_{Coeq(\Sigma_3)}\\
I&\longrightarrow^{\tau}&X&\rightarrow &X^{(2)}& \longrightarrow& X^{(3)}& \ldots\\
\end{array}
$$
Denote colimit of that diagram $(X,\tau)^{(\infty)}.$
Then we have canonical morphism $\zeta_{(X,\tau)}\colon (X,\tau)^{\infty} \to (X,\tau)^{(\infty)}.$
If $((X,\tau)^{(\infty)},\zeta_{(X,\tau)})$ is a coequalizer of $\eta(\Sigma_{\infty})$ then one can naturally call that pair an infinite symmetric product of $(X,\tau).$
UPDATE: Why it is non-trivial? Take a category with objects from the first diagram and terminal object. Then terminal object is colimit, but its automorphism group is trivial, therefore it does not contain $\Sigma_{\infty}.$
 A: This seems entirely straightforward unless I'm missing something.  For any $n\in\mathbb{N}$, $\Sigma_n$ acts on $X^{\otimes m}$ for any $m\geq n$ (on the first $n$ coordinates), and this action commutes with the maps in the colimit diagram.  Thus $\Sigma_n$ acts on the colimit of $X^{\otimes n}\to X^{\otimes (n+1)}\to\dots$, which is the same as $(X,\tau)^\infty$ because you've just taken a cofinal subdiagram.  Furthermore, these actions over different $n$ are compatible, so they glue together to give an action of $\Sigma_\infty$.
To put it another way, let $D$ be the category of diagrams in $C$ of the form $X_0\to X_1\to X_2\to\dots$, except that only the "tail" of the diagrams are required to be defined.  That is, for any $n\in\mathbb{N}$, a diagram of the form $X_n\to X_{n+1}\to X_{n+2}\to\dots$ also counts as an object of $D$, and a morphism between objects of $D$ need only be "eventually" defined (and parallel morphisms that eventually agree are identified).  It is straightforward to see that taking the colimit of the diagram gives a well-defined functor from $D$ to $C$.  Now observe that the diagram whose colimit is $(X,\tau)^\infty$ has an action of $\Sigma_\infty$ as an object of $D$, and hence so does its colimit.
