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