Let $p$ be prime and let $T^p$ be the $p$-torus and $\mathbb{Z}/p$ the cyclic group of order $p$ generated by $(12\ldots p)$. Consider the semidirect product $T^p\rtimes\mathbb{Z}/p$ with the natural action.
I would like to know if the representation ring of this semidirect product is known. Alternatively, what is known about the cohomology of the classifying space $H^*(B(T^p\rtimes\mathbb{Z}/p);G)$ for $G$ either $\mathbb{Q}$ or $\mathbb{Z}$?
For any finite dimensional representation, you can restrict to $T^p$ to get a decomposition into a direct sum of one-dimensional characters, which are given by $p$-tuples of integers. If a given character is diagonal (i.e., of the form $(n,\ldots,n)$), then it is a one-dimensional representation of $\mathbb{Z}/p$, i.e., characterized by a $p$-th root of unity. If a character is non-diagonal, then it is contained in the $p$-dimensional representation containing all cyclic permutations of that character, so the cyclic permutation class characterizes the non-diagonal irreducible representations.
Thus, we can describe the finite dimensional representations as finitely supported sheaves on the stack quotient $[\mathbb{Z}^p_{\mathbb{C}}/(\mathbb{Z}/p)]$. Tensor product is given by lifting to the cover and convolving (i.e., the cyclic symmetry on characters of $T^p$ is preserved).
As to the cohomology, here is an answer, which is probably an "over-kill". Probably you can find an answer in much older literature like Borel's.
Since $BT$ has torsion-free homology, COROLLARY 4.9 of JOHN ROBERT HUNTON The complex oriented cohomology of extended powers Annales de l’institut Fourier, tome 48, no 2 (1998), p. 517-534 applies.
With rational coefficients (you can definitely find this in Borel) you get just the invariants of $Q[x_1,\ldots x_p]$ under the action of cyclic groups. When $p=2$, the cyclic group being same as the permutation group, you get the ring of symmetric polynomials $Q[c_1,c_2]$ this way (which happens to be the cohomology of $BU(2)$, and of course, in this case your semi-direct product is the Weyl group of $U(2)$).
Things get a bit more complicated for integer coefficients, but you can read it off from the above corollary.
