There are already some questions with almost the same title, but they are more restrictive.

Let $G$ be a finite group, $H$ a subgroup, $V$ an irreducible representation of $H$, and $W=Ind_H^G V$ the induced representation.

Is there a simple way to compute the number of irreducible components of $W = Ind_H^G V$?

Let me precise my question by saying what kind of answers I would hope for. There is a well-known description of $Res_G^H Ind_H^G V = Res_G^H W$ as follows. Let $S$ be a set of representatives of the double classes of $H \backslash G / H$. For $s \in S$, let $H^s$ be the group $sHs^{-1} \cap H$ and $V^s$ the representation of $H^s$ on the space $V$ where $x \in H^s$ acts by $s^{-1}xs$ on $V$. Then $$Res_G^H W = \oplus_{s \in S} Ind_{H_s}^H V^s.$$ Using this description, one gets using Frobenius reciprocity twice that $$Hom_G (W,W) = Hom_H(V,Res_G^H W) = \oplus_{s \in S} Hom_H(V, Ind_{H^s}^H V^s) = \oplus_{s \in S} Hom_{H^s} (V,V^s)$$ which implies the well-known criterion of Mackey: $W$ is irreducible if and only if for all $s \in S$, $V$ and $V^s$ have no $H^s$-irreducible subrepresentations in common. So assuming we can compute $S$, the groups $H^s$, the representations $V^s$, and how $V$ and $V^s$ decomposes as representations of $H^s$ (this may be hard in practice, but let's assume we can do that), then we know when $W$ is irreducible, and in general we know how to determine the dimension of $Hom_G(W,W)$. (Of course all of this is standard cf. Serre Linear Representations for instance).

Yet this falls short of answering the question, because if for example you know that the dimension of $Hom_G(W,W)$ is $4$, that doesn't tell you if $W$ is the sum of 4 non-isomorphic irreducible representations, or of 2 copies of the same irreducible representation. What I'd like would be a way to tell us, by looking at the $H^s$ and the $V^s$ etc, of determine which one it is. Or more generally, how to "read" the decomposition of $W$ into irreducible reps. in terms of the $V^s$, $H^s$ etc. If it is not possible, I would also like an explanation why.