Let $G$ be a finite group and let $Rep(G)$ be its representation ring (as a group it is the free $\mathbb{Z}$-module on the irreducible complex reps). The collection of permutation representations $\mathbb{C}[\mathcal{O}]$ for $\mathcal{O}\cong G/H$ a $G$-orbit generate a $\mathbb{Z}$-subalgebra which we will denote by $Per(G) \subset Rep(G)$.

If $Per(G) = Rep(G)$ then it follows that all complex representations of $G$ are defined over $\mathbb{Z}$. To see this note that under our assumption for every representation $V$ there exists a representation $U$ defined over $\mathbb{Z}$ s.t. $U \oplus V = W$ is defined over $\mathbb{Z}$. We can now take $V_{\mathbb{Z}} = W_{\mathbb{Z}} / U_{\mathbb{Z}}$ as a $\mathbb{Z}$-form for $V$ (this is not really a precise proof, maybe a better arguemnt would be to explicitly write the projection operator which projectss onto the isotypic component of $V$ inside $\mathbb{Z}[G]$).

Is the converse true?

Question:Suppose every $\mathbb{C}$-representation of $G$ has a $\mathbb{Z}$-form, does it follow that $Per(G) = Rep(G)$? If not what's a counter example?

As an example when $G=S_n$ both of the properties are satisfied and this is in fact the only non-trivial example I know of.