Let $G$ be a group and $\mathcal G$ be a $\sigma$-algebra on $G$. Assume the inversion $G\ni x\mapsto x^{-1}$ is $(\mathcal G,\mathcal G)$-measurable and the group law $G^2\ni(x,y)\mapsto xy$$ is $(\mathcal G\otimes\mathcal G,\mathcal G)$-measurable.
Now let $$\tau_k:G^k\to G\;,\;\;\;x\mapsto x_1\cdots x_k$$ denote the iterated group law for $k\in\mathbb N$ and $\mu_i$ be a $\sigma$-finite measure on $(G,\mathcal G)$. Then the *convolution of $\mu_1,\ldots,\mu_k$ is defined to be the pushforward $\mu_1\ast\cdots\ast\mu_k:=\tau_k(\mu_1\otimes\cdots\otimes\mu_k)$ of $\mu_1\otimes\cdots\otimes\mu_k$ under $\tau_k$. If $\mu:=\mu_1=\cdots\mu_k$, we write $\mu^{\ast k}:=\mu_1\ast\cdots\ast\mu_k$.
I was able to show that $$\mu^{\ast2}(B)=\int\mu({\rm d}x)\mu(x^{-1}B)\;\;\;\text{for all }B\in\mathcal G\tag1,$$ where $x^{-1}B:=\{x^{-1}y:y\in B\}$. If we denote $$\tau_x:G\to G\;\;\;\;y\mapsto x^{-1}y$$ for $x\in G$, $$\kappa(x,B):=\mu(x^{-1}(B))\;\;\;\text{for }(x,B)\in(G,\mathcal E)$$ and the inverse element of $G$ by $1$, then $(1)$ is equivalent to $$\mu^{\ast2}(B)=\kappa^2(1,B)\;\;\;\text{for all }B\in\mathcal G\tag2,$$ where $\kappa^2$ denotes the composition of kernels. Are we able to generalize $(2)$ and show $$\mu^{\ast k}(B)=\kappa^k(1,B)\;\;\;\text{for all }B\in\mathcal G\tag3$$ for all $k\in\mathbb N$?