For two $n\times n$ hermitian matrices $A$, $B$, we have the trace inequality
$$\text{tr}(AB)\leq\sum_{i=1}^{n}\lambda_i(A)\lambda_i(B)$$
where the $\lambda_i(X)$ are the eigenvalues of X ordered in decreasing order.
Are there any inequalities similar to this but for unitary matrices? I am aware of the trace inequality for complex $n\times n$ matrices $A$ and $B$
$$\text{tr}(AB)\leq\sum_{i=1}^{n}\sigma_i(A)\sigma_i(B)$$
where the $\sigma_i(X)$ are singular values of X ordered in decreasing order. For unitary matrices this yields a very uninteresting bound, as all singular values are 1.
I don't think you can improve the bound for arbitrary unitary matrices. Indeed, you can just take $B=A^*$, the hermitian conjugate of $A$ (also its inverse) to see that you get equality. Maybe a more trivial example is to take $A=B=I$
Another way to see it is to realize that for any two matrices $M$ and $N$, $$\left|\text{tr}(MN^*)\right| \leq \sqrt{\text{tr}(MM^*)}\sqrt{\text{tr}(NN^*)}\tag{1}$$ This is the Cauchy-Schwarz inequality for the hermitian product defined by $(M, N) \mapsto \text{tr}(MN^*)$ on matrices. Here, $N^*$ is the hermitian conjugate of $N$.
Now, if you take $A$ and $B$ two unitary matrices, then applying the above with $M=A$ and $N=B^*$ yields $$\left|\text{tr}(AB)\right| \leq \sqrt{\text{tr}(I)}\sqrt{\text{tr}(I)}=n$$ which is the same as your inequality. Now we know that the Cauchy-Schwarz inequality is sharp, that is, it cannot be improved upon without losing its generality (i.e. if you don't consider subsets of all matrices, even unitary ones, since equality is achieved easily as demonstrated above).