Let $K, K'$ be knots in $S^3$, and $T, T'$ the boundaries of their tubular neighborhoods. 


Recall that by theorems of Waldhausen, and Gordon and Luecke, one knows the following: an isomorphism $[\pi\_1(T) \to \pi\_1(S^3 \setminus K)] \cong [(\pi\_1(T') \to \pi\_1(S^3 \setminus K')]$ implies that $K$ and $K'$ are isotopic. 



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Note: In a previous version of this question, I asked whether this is possible if you interpret the category above as a tensor category; in the comments it is pointed out that then, yes, at least if you allow big representations. But, now I realize that in fact I don't have the tensor product structure. 



What does the representation category of the knot group know?
So long as I allow myself infinite size representations, and writing $\mathbb{Z}[G]-mod$ for representations of $G$ in $\mathbb{Z}$-modules, then so long as I have the forgetful functor


$$\mathbb{Z}[G]-mod \to \mathbb{Z}-mod$$


I can recover $\mathbb{Z}[G]$ as endomorphisms of this functor. 


In general, one cannot recover $G$ even from $\mathbb{Z}[G]$, [even for finite groups](http://www.jstor.org/stable/3062112) $G$. 


However, as it turns out, knot groups are known to be left orderable, and it is known that for left orderable groups, the only units in $\mathbb{Z}[G]$ are the obvious ones, from which $G$ can be recovered. 


