OK with a few pointers from that paper I've got a proof that \\(\lambda_I = \rho_I : I\otimes I\to I\\).

First we note that \\(\rho\\) is natural, so for any \\(\phi : I\otimes I\to I\\) we have \\(\phi\circ\rho_{I\otimes I} = \rho_I\circ(\phi\otimes 1_I)\\).

Setting \\(\phi = \rho_I\\) gives us \\(\rho_I\circ\rho_{I\otimes I} = \rho_I\circ(\rho_I\otimes 1_I)\\).

But \\(\rho_I\\) is an isomorphism, so we can conclude that \\(\rho_{I\otimes I} = \rho_I\otimes 1_I\\).

Setting \\(\phi = \lambda_I\\) gives us \\(\lambda_I\circ\rho_{I\otimes I} = \rho_I\circ(\lambda_I\otimes 1_I)\\).

Combining these two gives us \\(\lambda_I\circ(\rho_I\otimes 1_I) = \rho_I\circ(\lambda_I\otimes 1_I)\\).

Here's a couple of diagrams for this stage:

For the second bit we need two "triangle" equations:

\\(\qquad\rho_A\otimes 1_C = (1_A\otimes\lambda_C)\circ\alpha_{A, I, C} : (A\otimes I)\otimes C\to A\otimes C\\)

\\(\qquad\lambda_B\otimes 1_C = \lambda_{B\otimes C}\circ\alpha_{I, B, C} : (I\otimes B)\otimes C\to B\otimes C\\)

The first of these is one of the standard coherence equations from the definition of a monoidal category. The second one can be derived from the first one together with the "pentagon" diagram – although this is a little fiddly, so let's just assume it for now.

Setting \\(A = B = C = I\\) in the triangle equations, and using the naturality of \\(\lambda\\), we get the following diagram:

The outer square tells us \\(\lambda_I\circ(\lambda_I\otimes 1_I) = \lambda_I\circ(\rho_I\otimes 1_I)\\).

But \\(\lambda_I\\) is an isomorphism, hence \\(\lambda_I\otimes 1_I = \rho_I\otimes 1_I\\).

Combining this with \\(\lambda_I\circ(\rho_I\otimes 1_I) = \rho_I\circ(\lambda_I\otimes 1_I)\\) proved above gives us \\(\lambda_I = \rho_I\\). QED