Khovanov's Frobenius algebra (used in the definition of Khovanov homology) is $\mathbb{Z}[X]/X^2$ with the comultiplication. $\Delta(X)=X\otimes X, \Delta(1)=1\otimes X+X\otimes 1$ and the trace $\epsilon(1)=0, \epsilon(X)=1$. (Although it is a bit more general but for my question this suffices.) The definition of a "self dual Frobenius extension" is given in http://arxiv.org/abs/math/0411447 and it is basically an isomorphism of the algebra with its dual under which multiplication is sent to comultiplication and the unit to the trace. This is easily seen to be the case for the Khovanov algebra with the isomorphism that sends $1$ to $X$ and vice-verse.

Now my question is whether $\Delta$ is sent to the multiplication $m$ under this isomorphism. $\Delta^*$ sends $1\otimes X+X\otimes 1 \to X$, $1\otimes 1 \to 1$ and $X\otimes X \to 0$ but this is not exactly $m$.