 Solution

For the Category of sets, take the monoidal tensor product to be the Cartesian product, $A\otimes B=A\times B$, with unit a singleton $1$. On morphisms use $(f\times g)(a,b)=(f(a),g(b))$. Define the associator by $((a,b),c)\mapsto (a,(b,c))$ and the unitors by $(\ast ,a)\mapsto a$ and $(a,\ast )\mapsto a$. These are natural isomorphisms, with inverses obtained by rebracketing or adjoining $\ast$. Both paths in the pentagon send $(((a,b),c),d)$ to $(a,(b,(c,d)))$, and both paths in the triangle send $((a,\ast ),b)$ to $(a,b)$. This proves the axioms elementwise.

There is a second structure: use disjoint union $A⨿B$ and unit $\varnothing$. Define the associator by preserving each element and its summand while changing the nested tags, and define the unitors by removing the empty summand. These are natural and invertible. In the pentagon, either path takes an element in any of the four summands to that same element in the same summand of $A⨿(B⨿(C⨿D))$; the triangle removes an empty intermediate summand on either path. Thus both axioms hold.

These product and coproduct monoidal structures on sets have different units, $1$ and $\varnothing$, which are not isomorphic. They are even inequivalent as monoidal categories: an equivalence of the underlying Category of sets preserves terminal objects, so it cannot take the singleton unit of the product structure to the empty unit of the coproduct structure. Hence the monoidal structure on sets is not unique.