How do I know if an irreducible representation is a permutation representation?
The map from the Burnside ring to the representation ring is well studied. Andreas Dress wrote many papers on this in the 1970's. See also work of Tammo tom Dieck, who was applying things to equivariant topology.
There is a detecting character ring for each of these. Characters for elements of the Burnside ring are integer valued functions from conjugacy classes of subgroups of $G$, and an evident map between the two character rings. A classic character $\chi$ that is integer valued is in the image of this map exactly when $\chi(g) = \chi(h)$ if $g$ and $h$ generate conjugate cyclic groups.
In a 1987 paper in Comm. Math. Helv., Peter Webb gives an explicit basis for the kernel of the map from the Burnside ring to the Representation ring, after inverting the order of $G$.
This should get you going, if you want to learn about this.
First of all, note that $S_6$ has a doubly transitive action on 10 points, obtained from the action of $PSL_2(9)=A_6$ on the projective line over $\mathbb{F}_9$ by adding the Galois automorphism of $\mathbb{F}_9$, and this is the only faithful permutation action of $S_6$ of degree 10. $S_6$ has two irreducible characters of degree 9, one obtained from the other by tensoring with the alternating degree 1 charater. Only one of them, say $\chi$, can come up in the doubly transitive permutation character, as all the values $\chi$ are bounded from below by $-1$.
With GAP, it is very easy to answer everything on $S_6$. (One can replace this by an more human argument; you can also look up the character table in question in the literature, e.g. in the "Atlas of Finite Simple Groups".) In particular you can compute the irreducible characters of degree 9:
[ 9, -3, 1, -3, 0, 0, 0, 1, 1, -1, 0 ]
[ 9, 3, 1, 3, 0, 0, 0, -1, 1, -1, 0 ]
Thus $\chi$ is the latter one, and it does occur in the doubly transitive degree 9 permutation character of $S_6$ (add 1 to each entry).
To give you an example that having entries at least $-1$ is not enough to get a constituent of a doubly transitive character, note that $M_{11}$ (the smallest Mathieu sporadic simple group) has degrees 44 and 55 $\mathbb{Z}$-valued irreducible characters with values at least $-1$, but no degree 45 and 56 doubly transitive permutation representations (for case of 56, the group order is not divisible by 56).