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 $\Sigma_{\infty}.$**