Let a group $G$ act on a finite set $\Omega$. Suppose that the corresponding permutation character has a regular component. Does it follow that $\Omega$ has a regular $G$orbit? (The converse is obviously true.)
No, this is false in general, and the smallest counterexample is the Klein 4group $G=C_2\times C_2$. It can act transitively on a set of order 2 in three different ways, with 3 different $C_2$'s in the kernel. Call these sets $\Omega_1, \Omega_2, \Omega_3$, and their permutation characters $1+\chi_1, 1+\chi_2, 1+\chi_3$. Then the character of $\Omega=\Omega_1\coprod\Omega_2\coprod\Omega_3$ is $3+\chi_1+\chi_2+\chi_3$, which is 1+1+regular character of $G$, but $\Omega$ does not have a regular orbit. 


Not every noncyclic finite group has such a relation. E.g., the quaternion group of order 8 doesn't. (I meant this as a comment on Alex's answer, but I don't have enough reputation to comment.) 


No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$. One systematic source of such counterexamples are Brauer relations in which the regular set $G/1$ enters with nonzero coefficient, of which both Tim's and my answer are particular examples (for more on Brauer relations see e.g. this MO question, as well as google). 

