According to Wikipedia,
This result is known as Cramér's decomposition theorem, and is equivalent to saying that the convolution of two distributions is normal if and only if both are normal.
The context isn't that important, but basically that statements implies that repeated convolution with a normal distribution is equivalent to a normal distribution itself. However, there is no statement on the connection between the pre- and post-parameters, so my questions are:
Given a normal distribution $$n(x;\mu,\sigma) = \frac1{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ with mean $\mu$ and deviation $\sigma$, what parameters achieve the same (single) convolution $n(x;\mu',\sigma')$ that performing the original convolution twice would yield, i.e. what $\mu', \sigma'$ satisfy $$n(x;\mu,\sigma)\ast n(x;\mu,\sigma)\ast f(s) = n(x;\mu',\sigma')\ast f(x) \quad(\forall f(x))$$
Assuming the result is not simply $\mu'=\mu$ and $\sigma'=\sigma$, does repeating this re-parametrization $\mu\to\mu'$ and $\sigma\to\sigma'$ ultimately converge?
Let's tackle this using the (symmetric) Fourier transform
$$\mathcal F\{f(x)\}(k) := \hat f(k) := \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-ikx}$$
such that
$$\hat n(k;\mu,\sigma) = \frac1{\sqrt{2\pi}}e^{i\mu k} e^{-\frac{\sigma^2}2 k^2}$$
and use the convolution theorem:
$$\begin{align*} \mathcal F\{a\ast b\} &= \mathcal F\{a\} \cdot \mathcal F\{b\} \\\Rightarrow \mathcal F\{n(x;\mu,\sigma)\ast n(x;\nu,\rho)\}(k) &= \frac1{\sqrt{2\pi}}e^{i(\mu+\nu)k}e^{-\frac{\sigma^2+\rho^2}2k^2} \\ &= \hat n(k; \mu+\nu, \sqrt{\sigma^2+\rho^2}) \end{align*}$$
i.e. for $\nu=\mu$ and $\rho=\sigma$ we have
$$\begin{align} \mu &\to 2\mu, \\\sigma &\to \sqrt2 \sigma. \end{align}$$
So aside from the centered Delta-distribution ($\mu=0=\sigma$) the "convergence" is towards a distribution where each value is equally probable (but due to it stretching over all $\mathbb R$, that is almost 0).