Different induced representations - same simples? is the following case possible: 
$\pi_1, \pi_2$ two simple representations of the same subgroup over an arbitrary field. $\operatorname{Ind}(\pi_1)$ and $\operatorname{Ind}(\pi_2)$ have equal components even if $\pi_1$ and $\pi_2$ are different.
 A: The case you asked has easy examples where the induced representations are isomorphic (take the two non-trivial irreducible representations of the alternating group of degree 3 and induce them to the symmetric group of degree 3; this works over any field containing a primitive 3rd root of unity).
I suspect you wanted an example where the induced representations are not isomorphic, but where their composition factors (or equivalently their Brauer characters) are equal. Such examples exist: 
Let $G = \operatorname{GL}(2,3)$ and let $H$ be a Borel subgroup (normalizer of a Sylow 3-subgroup). Both $\pi_i$ are 1-dimensional representations of $H$ over the field of 3 elements, and so are determined by their values on the diagonal matrices $\left[\begin{smallmatrix} a & 0 \\ 0 & b \end{smallmatrix}\right]$. Let $\pi_1(a,b) = a$ and $\pi_2(a,b)=b$. Then the induced representations clearly have the same Brauer characters. However, these modules are non-isomorphic (the socle of one is the head of the other; each is 2-dimensional and self-dual so that its head and socle are duals; the whole module is 4-dimensional).
