The conjecture is true and holds, in a generalized version, whenever $K$ is an abelian regular subgroup. In fact by Theorem 1.9(d) in O. Tamaschke, *Zur Theorie der Permutationsgruppen mit regulärer Untergruppe. I*, Math. Zeit. (1963) **80** 328–354, there is an even more general result that holds when $K$ is not necessarily abelian. A special case of this theorem is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle_\mathbb{C}$ has the structure of a unitary central $S$-ring. In particular, it is closed under the natural product.

Below I'll give a proof in the cyclic case, using some ideas from Wolfgang Knapp *On Burnside's Method*, J. Alg. (1995) **175** 644–660. In outline, what we show is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle$ under multiplication is isomorphic to the Schur ring for $G$ with respect to the product that sends two $H$-invariant subsets to their intersection.

*Setup.* Let $\zeta \in \mathbb{C}$ be a primitive $d$-th root of unity. For $j \in \{0,\ldots, d-1\}$ let

$$v_j = \sum_{i=0}^{d-1} \zeta^{-ij} k^i \in \mathbb{C}K.$$

The $v_j/d$ are the orthogonal primitive idempotents in the commutative group algebra $\mathbb{C}K$. Since $v_j k = \zeta^j v_j$, the subspace $\langle v_j \rangle$ affords the character $\theta^j$. Let $\odot$ denote the product on $\mathbb{C}K$ defined by multiplying coefficients of each $k^j$: thus if $d=3$ then

$$(a_0+a_1k+a_2k^2) \odot(b_0+b_1k+b_2k^2) = a_0b_0 + a_1b_1 k + a_2b_2k^2.$$

Let $\mathcal{C}(K)$ denote the character ring of $K$. Consider the linear map $\mathcal{C}(K) \rightarrow \mathbb{C}K$

$$\chi \mapsto \sum_{i=0}^{d-1} \overline{\chi(k^j)}k^j.$$

Since the image of $\theta^j$ is $v_j$, this map is a linear isomorphism. Observe that

$$v_j \odot v_{j'} = v_{j + j' \,\mathrm{mod}\; d}.$$

Therefore $\mathcal{C}(K)$, with its usual product, corresponding to the tensor product of modules, is isomorphic to $(\mathbb{C}K, \odot)$.

*Proof of conjecture.* Let $G$ act on $\{0,1,\ldots,d-1\}$. Let $H \le G$ be the point stabiliser of $0$. Let $\mathcal{O}_0, \mathcal{O}_1, \ldots, \mathcal{O}_\ell$ be the orbits of $H$ on $\{0,1,\ldots,d-1\}$, ordered so that $\mathcal{O}_0 = \{0\}$. (Note that the number of orbits is the rank of the $G$-action, which is the number of irreducible constituents of $\pi$, namely $\ell+1$.) Identify the $G$-set $\{0,1,\ldots,d-1\}$ with the regular subgroup $K$ by $j \mapsto k^j$. The orbit sums are then

$$e_m = \sum_{j \in \mathcal{O}_m} k^j \in \mathbb{C}K$$

for $m \in \{0,1,\ldots,\ell\}$. By definition $S = \langle e_0, e_1, \ldots, e_\ell \rangle_\mathbb{C}$ is the Schur ring for $G$. The $e_m$ are Schur's `primitive elements' and are primitive idempotents for the $\odot$ product. (The Schur ring is also a unital ring under the normal product on $\mathbb{C}K$, but, perhaps surprisingly, we do not need this fact here.)

By our identification, $\mathbb{C}K$ affords the natural permutation module for $G$. Let

$$\mathbb{C}K = V_0 \oplus V_1 \oplus \cdots \oplus V_\ell$$

be its direct sum decomposition into irreducible representations, where

$$V_0 = \langle 1_K + k + \cdots + k^{d-1} \rangle$$

affords the trivial representation, and $V_i$ affords the character $\pi_i$ in the question. Note that $V_0 = \langle v_0 \rangle$. More generally, let $B_i$ be the set of $j$ such that $\langle \pi_i\!\!\downarrow_K, \theta^j \rangle = 1$. Then $V_i = \langle v_j : j \in B_i\rangle$. Moreover, by Frobenius reciprocity

$$\langle \pi_i \!\!\downarrow_H, \mathbb{1}_H \rangle = \langle \pi_i, \mathbb{1}_H\!\!\uparrow^G \rangle = \langle \pi_i, \pi \rangle = 1$$

for each $i$, so each $V_i$ has a unique (up to scalars) $H$-invariant vector. Since $e_0 = \frac{1}{d}(v_0 + v_1 + \cdots + v_{d-1})$ is $H$-invariant, taking the projection into $V_i$ we see that a suitable choice for this $H$-invariant vector is $\sum_{j \in B_i} v_j$. Therefore an alternative basis for the Schur ring $S$ is

$$\big\{ \sum_{j \in B_i} v_j : j \in \{0,1,\ldots, d-1\bigr\}.$$

Under the isomorphism in the 'setup', we have

$$\phi_i = \sum_{j \in B_i} \theta^j \mapsto \sum_{j \in B_i} v_j.$$

Therefore, by the inverse of this isomorphism, $(S, \odot)$ is isomorphic to $\langle \mathbb{1}_K, \theta_1, \ldots, \theta_\ell \rangle$. In particular, the linear span of characters is closed under the product in $\mathcal{C}(K)$. $\Box$