Let $E$ be a $\mathbb R$-Banach space, $\mathcal M_1(E)$ (resp. $\mathcal M_1^\infty(E)$) denote the set of probability measures (resp. infinitely divisible probability measures) on $E$, $\varphi_\mu$ denote the characteristic function of $\mu\in\mathcal M_1(E)$, $$\mathcal C_1(E):=\left\{\varphi_\mu:\mu\in\mathcal M_1(E)\right\}$$ and $$\mathcal C_1^\infty(E):=\left\{\varphi_\mu:\mu\in\mathcal M_1^\infty(E)\right\}.$$

Remember that $\mathcal M_1(E)$ is infinitely divisible, i.e. $\mu\in\mathcal M_1^\infty(E)$, if and only if $$\forall n\in\mathbb N:\exists\nu\in\mathcal M_1(E):\mu=\nu^{\ast k}\tag1$$ or, equivalently, $$\forall n\in\mathbb N:\exists\psi\in\mathcal C_1(E):\varphi_\mu=\psi^n\tag2.$$

The first question is whether $\nu$ (resp. $\psi$) in $(1)$ (resp. $(2)$) are unique (if they exist). The second question is whether, given $\mu\in\mathcal M_1^\infty(E)$, there is a unique continuous convolution semigroup $(\mu_t)_{t\ge0}$ on $E$ with $\mu_1=\mu$.

For both questions, I'm only able to give a positive result when $E=\mathbb R^d$ for some $d\in\mathbb N$ and I would really like to know whether there are generalizations.

Assuming $E=\mathbb R^d$ for some $d\in\mathbb N$, we are able to show that for every $\varphi\in C^0(\mathbb R^d,\mathbb C\setminus\{0\})$ with $\varphi(0)=1$, there is a unique $f\in C^0(\mathbb R^d,\mathbb C)$ with $f(0)=0$ and $\varphi=e^f$. Moreover, for every $k\in\mathbb N$, there is a unique $g\in\mathbb C^0(\mathbb R^d,\mathbb C)$ with $\varphi=g^k$; in fact, $g=e^{f/k}$. This can be applied to every $\varphi\in C_1(\mathbb R^d)$ and hence we not only obtain uniqueness of $\nu$ (resp. $\psi$) in $(1)$ (resp. $(2)$), but even that $\nu$ (resp. $\psi$) are infinitely divisible as well.

By this result it's easy to see that, for every $m,n\in\mathbb N$, there is a unique $\mu_{m/n}\in\mathcal M_1(\mathbb R^d)$ with $\mu^{\ast m}=\mu_{m/n}^{\ast n}$; in fact, $\mu_{m/n}=\mu_{1/n}^{\ast m}$. If $t\ge0$, there is a $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)\cap\mathbb Q$ with $t_n\xrightarrow{n\to\infty}t$ and hence $$\varphi_{\mu_{t_n}}=e^{t_nf}\xrightarrow{n\to\infty}e^{tf}\tag3,$$ where $f$ is as above, but corresponding to the choice $\varphi=\varphi_\mu$ for our given $\mu$. And here it seems like that we again need that $E=\mathbb R^d$, since only then Lèvy's continuity theorem is applicable and yields the existence of a unique $\mu_t\in\mathcal M_1(\mathbb R^d)$ with $\varphi_{\mu_t}=e^{tf}$. Once again, it's easy to see that $\mu_t$ is infinitely divisible as well.