Quick summary and notation. Provided we have some measure $\mu$ on the Borel space $\mathscr{B}(\mathbb{R}^d)$, one can define an $n$-fold convolution measure $\mu \ast \cdots \ast \mu$ on $\mathscr{B}(\mathbb{R}^d)$ by evaluating each $A \in \mathscr{B}(\mathbb{R}^d)$ as follows. $$ \big(\mu \ast \cdots \ast \mu\big)(A) = \int_{\mathbb{R^d}}\cdots \int_{\mathbb{R^d}} 1_A(x_1 + \cdots + x_n) \mu({\rm d}x_1) \cdots \mu({\rm d}x_n) $$ By Fubini, this object $\mu_n = \mu \ast \cdots \ast \mu$ too is a measure, and its characteristic function $\hat\mu_n$ has a very nice form in terms of that $\hat\mu$ of $\mu$. $$\begin{aligned} \hat\mu_n(u) &= \int_{\mathbb{R}^d} \exp\langle iu,x\rangle \mu_n({\rm d}x) \\ &= \int_{\mathbb{R}^d} \cdots \int_{\mathbb{R}^d}\exp\langle iu, x_1 + \cdots + x_n\rangle \mu({\rm d}x_1) \cdots \mu({\rm d}x_n) \\ &= \bigg( \int_{\mathbb{R}^d} \exp\langle iu, x \rangle \mu({\rm d}x) \bigg)^n \\ &= \big(\mu(u)\big)^n \end{aligned}$$
This is a nice homomorphism between measures and their characteristic functions. $$ \widehat{\mu \ast \mu} = \hat\mu \cdot \hat\mu$$
My question. Provided some measure $\mu$ and positive real number $r > 0$, can I guarantee some measure $\mu_r$ with characteristic function being the corresponding exponential? $$ \hat\mu_r(u) = \big(\mu(u)\big)^r $$
Surely, there are some cases in which this is obvious, like $\mu$ being degenerate or belonging to some nice exponential family of probability measures. I simply wonder if there is a general result for all (probability) measures $\mu$. Also, it seems like the comprehensive theory of Fourier transforms on the Schwartz class and tempered distributions might lend itself somewhere, but my knowledge is very weak in this area.
No. Setting $r = \frac{1}{n}$ this would imply that $\mu$ is itself the $n$-fold convolution of some other measure $\mu_{\frac{1}{n}}$; for probability measures this property is called infinite divisibility and it generally does not hold. For example, the uniform distribution on $[0, 1]$ is not infinitely divisible; the Wikipedia article says this follows from the fact that it has bounded support (but I don't know off the top of my head why this is enough).