Is $(G\rtimes H,H)$ a Gelfand pair iff $G$ is abelian? Let $G$ be a finite group and $K \subset G$ a subgroup.  Then $(G,K)$ is a Gelfand pair if the double coset Hecke algebra $\mathbb{C}(K \backslash G /  K)$ is commutative.
Let $H$ be a subgroup of $Aut(G)$.  We will consider $H$ and $G$ as subgroups of $G\rtimes H$.
Lemma: $(G\rtimes H,H)$ is a Gelfand pair  iff $\forall g,g' \in G$, $HgHg'H = Hg'HgH$.
Proposition: If $G$ is abelian, then $(G\rtimes H,H)$ is a Gelfand pair.
proof: $\forall h_1, h_2, h_3 \in H$ and $\forall g,g' \in G$, we have:
$$h_1gh_2g'h_3 = h_1 g \sigma_{h_2}(g')  h_2 h_3 =  h_1 \sigma_{h_2}(g') g   h_2 h_3 = h_1 h_2 g' h_2^{-1} g h_2h_3$$ which means that $HgHg'H \subseteq Hg'HgH$. Idem, $Hg'HgH \subseteq HgHg'H$. $\square$
Question: Is the converse true [i.e. $(G\rtimes H,H)$  Gelfand pair $\Rightarrow$ $G$  abelian]? 
 A: It is perhaps worth pointing out the special case that if the semidirect product $GH$ is a Frobenius group with kernel $G$ and complement $H$, then $(GH,H)$ is a Gelfand pair if and only if $G$ is Abelian.
In general, we have a Gelfand pair from such a semidirect product if and only if the permutation character $\theta = {\rm Ind}_{H}^{GH}(1)$ is multiplicity free. One explanation why we always get a Gelfand pair when $G$ is Abelian is that even the restriction ${\rm Res}^{GH}_{G}(\theta)$ is the regular character of $G$, which is multiplicity free when $G$ is Abelian.
Suppose now that $GH$ is such a Frobenius group, but that $G$ is non-Abelian. Then $G$ has a non-linear irreducible character $\mu$, which occurs with multiplicity $\mu(1)$ in ${\rm Res}^{GH}_{G}(\theta)$. On the the other hand,
there is just one irreducible character $\chi$ of $G$ which contains $\mu$ with non-zero multiplicity on restriction to $G$, and that is $\chi = {\rm Ind}^{GH}_{G}(\mu)$. Hence $\chi$ must occur with multiplicity $\mu(1) > 1$ in $\theta $, so that $(GH,H)$ is not a Gelfand pair.
More generally, when $H$ is cyclic, or when $H$ is Abelian of order relatively prime to $|G|$, it is possible to use Clifford's theorem to check that $(GH,H)$ is not a Gelfand pair if there is an irreducible character $\mu$ of $G$ such that $\mu(1) > |I_{H}(\mu)|$, where $I_{H}(\mu) = \{h \in H : \mu^{h} = \mu \}$, the Frobenius group case being an extreme example of this where $I_{H}(\mu)$ is the trivial group.
A: No, let $G$ be an arbitrary group and take $H=G$ acting on itself by conjugation. Then $H$ becomes the diagonal in $G\rtimes H\cong G\times G$. This is well known to be a Gelfand pair.
PS: The Hecke algebra of the pair $(G\rtimes H,H)$ is in fact equal to the fixed point algebra $\mathbb C[G]^H$. This explains your observation when $G$ is abelian.
