An example in which there are two isomorphic irreducible modules in the decomposition is the group ${\rm PSL}(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.

Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.

```
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
```