What is the most surprising characterization of the Gaussian (normal) distribution? A standardized Gaussian distribution on $\mathbb{R}$ can be defined by giving explicitly its density: 
$$ \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$
or its characteristic function. 
As recalled in this question it is also the only distribution for which the sample mean and variance are independent. 
What are other surprising alternative characterization of Gaussian measures that you know ? I will accept the most surprising answer 
 A: Let $\eta$ and $\xi$ be two independent random variables with a common symmetric distribution such that 
$$ P\left ( \left |\frac{\xi+\eta}{\sqrt{2}}\right | \geq t \right )\leq P(|\xi|\geq t).$$
Then these random variables are gaussian. (Obviously, if the $\xi$ and $\eta$ are centered gaussian, it is true.) 
This is the Bobkov-Houdre Theorem 
A: Suppose one is estimating a location parameter using i.i.d. data $\{x_1,...,x_n\}$. If $\bar{x}$ is the maximum likelihood estimator, then the sampling distribution is Gaussian. According to Jaynes's Probability Theory: The Logic of Science pp. 202-4, this was how Gauss originally derived it. 
A: A more particular characterisation of the normal distribution among the class of infinitely divisible distributions is presented in Steutel and Van Harn (2004).

A non-degenerate infinitely divisible random variable $X$ has a normal distribution if and only if it satisfies
  $$-\limsup_{x\rightarrow\infty}\dfrac{\log{\mathbb P}(\vert X\vert>x)}{x\log(x)}=\infty.$$ 

This result characterises the normal distribution in terms of its tail behaviour.
A: In the context of image smoothing (e.g. scale space), the Gaussian is the only rotationally symmetric separable* kernel.
That is, if we require
$$F[x,y]=f[x]f[y]$$
where $[x,y]=r[\cos\theta,\sin\theta]$, then rotational symmetry requires
\begin{align}
F_\theta &= f'[x]f[y]x_\theta+f[x]f'[y]y_\theta \\
 &= -f'[x]f[y]y+f[x]f'[y]x = 0 \\
&\implies \\
\frac{f'[x]}{xf[x]} &= \frac{f'[y]}{yf[y]} = \mathrm{const.}
\end{align}
which is equivalent to $\log\big[f[x]\big]'=cx$.
Requiring that $f[x]$ be a proper kernel then requires the constant be negative and the initial value positive, yielding the Gaussian kernel.

*In the context of probability distributions, separable means independent, while in the context of image filtering it allows the 2D convolution to be reduced computationally to two 1D convolutions.
A: My personal most surprising is the one about the sample mean and variance, but here is another (maybe) surprising characterization: if $X$ and $Y$ are IID with finite variance with $X+Y$ and $X-Y$ independent, then $X$ and $Y$ are normal.
Intuitively, we can usually identify when variables are not independent with a scatterplot.  So imagine a scatterplot of $(X,Y)$ pairs that looks independent.   Now rotate by 45 degrees and look again: if it still looks independent, then the $X$ and $Y$ coordinates individually must be normal (this is all speaking loosely, of course).
To see why the intuitive bit works, take a look at
$$
\left[
\begin{array}{cc}
\cos45^{\circ} & -\sin45^{\circ} \newline 
\sin45^{\circ} & \cos45^{\circ} 
\end{array}
\right]
\left[
\begin{array}{c}
x \newline
y
\end{array}
\right]= \frac{1}{\sqrt{2}}
\left[
\begin{array}{c}
x-y \newline
x+y
\end{array}
\right]
$$
A: Recently Ejsmont [1] published article with new characterization of Gaussian: 
Let $(X_1,\dots, X_m,Y) \textrm{ and } (X_{m+1},\dots,X_n,Z)$ be independent  random vectors with all moments, where  $X_i$ are
nondegenerate,  and let statistic $\sum_{i=1}^na_iX_i+Y+Z$
have a distribution which depends only on $\sum_{i=1}^n a_i^2$, where $a_i\in \mathbb{R}$ and $1\leq m < n$.  Then $X_i $ are independent and have the same normal distribution with zero means and $cov(X_i,Y)=cov(X_i,Z)=0$ for $i\in\{1,\dots,n\}$. 
[1]. Ejsmont, Wiktor. "A characterization of the normal distribution by the independence of a pair of random vectors." Statistics & Probability Letters 114 (2016): 1-5.
A: The continuous distribution with fixed variance which maximizes differential entropy is the Gaussian distribution.
A: There's an entire book written about this: "Characterizations of the normal probability law", A. M. Mathai & G. Perderzoli.  A brief review in JASA (Dec. 1978) mentions the following:

Let $X_1, \ldots, X_n$ be independent random variables.  Then $\sum_{i=1}^n{a_i x_i}$ and $\sum_{i=1}^n{b_i x_i}$ are independent, where $a_i b_i \ne 0$, if and only if $X_i$ [are] normally distributed.

A: Stein’s Lemma provides a very useful characterization.  $Z$ is standard Gaussian iff 
$$E f’(Z) = E Z f(Z)$$
for all absolutely continuous functions $f$ with $E|f’(Z)| < \infty$.
A: Gaussian distributions are the only sum-stable distributions with finite variance. 
A: Theorem [Herschel-Maxwell]: Let $Z \in \mathbb{R}^n$ be a random vector for which (i) projections into orthogonal subspaces are independent and (ii) the distribution of $Z$ depends only on the length $\|Z\|$. Then $Z$ is normally distributed.
Cited by George Cobb in Teaching statistics: Some important tensions (Chilean J. Statistics Vol. 2, No. 1, April 2011) at p. 54.
Cobb uses this characterization as a starting point for deriving the $\chi^2$, $t$, and $F$ distributions, without using Calculus (or much probability theory).
A: This is not a characterization but a conjecture, which dates back from 1917 and is due to Cantelli: 

If $f$ is a positive function on $\mathbb{R}$ and $X$ and $Y$ are $N(0,1)$ independent random variables such that $X+f(X)Y$ is normal, then $f$ is a constant almost everywhere.

Mentioned by Gérard Letac here.
A: Its characteristic function has the same form as its pdf.  I am not sure of another distribution which does that. 
A: The expectation plus minus the standard deviation are the saddle points of the function. 
