If the sum, min, max of iid rvs has the same distribution as the rvs, is it sufficient? It seems intuitive. For example, the $Min(x_1,x_2...x_n)$ where the $x_i's$ are iid $Geo(\theta)$ is also $Geo(\theta)$. This means that the Min, is also a sufficient stat for $\theta$, right?
 A: $$
\DeclareMathOperator{geo}{Geo}
\DeclareMathOperator{P}{\mathbb{P}}
$$
Suppose $X_i\sim \geo(\theta)$ iid for $i\in[n]$. Then:
$$\P\{\min_{i\in[n]}X_i> k\}=\prod_{i=1}^n\P\{X_i> k\}=(1-\theta)^{kn}$$
But notice that this implies $\P\{\min_iX_i\le k\}=1-(1-\theta)^{nk}$, where this is exactly the CDF of a $\geo(1-(1-\theta)^n)$-distributed rv, not a $\geo(\theta)$ one. Intuitively, the minimum of $n$ equally distributed, independent things won't usually be distributed the same way, since it's smaller than most of those $n$ things all the time.
Assuming $X_i\sim \geo(\theta)$ are iid for some unknown $\theta$, $\min_i X_i$ is not a sufficient statistic for $\theta$. It suffices to show this for $n=2$. $\P\{X_1=x_1,X_2=x_2\}=(1-\theta)^{x_1+x_2-2}\theta^2$, but this cannot be factored into a product of a function of $x_1,x_2$ alone and $\theta,\min(x_1,x_2)$ alone (take logarithms, the derivative must only be a function of the latter two, but it isn't).
In general, having a statistic distributed the same way as a rv characterized by a parameterized distribution doesn't mean anything. What matters is that the functional relationship from the rv to the statistic suffices to describe the relationship between the parameter and rv; more precisely, the parameter and the random variable are conditionally independent given the statistic.
