If $ X\sim Beta(\alpha,\beta)$, then $f(x;\alpha,\beta) = {\Gamma(\alpha+\beta) \over \Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} $
and $ f(\underline x;\alpha,\beta) = ({\Gamma(\alpha+\beta) \over \Gamma(\alpha)\Gamma(\beta)})^{n}(\prod x_{i})^{\alpha-1}(\prod (1-x_{i}))^{\beta-1}$
Thus, the sufficient statisitcs is $(\prod x_{i},\prod (1-x_{i}))$ by Fisher–Neyman factorization theorem.
Here $ h(\underline x) = 1$.
If $ \alpha = \beta $, then sufficient statistics is $(\prod x_i(1-x_i))$. Again $h(\underline x) = 1$
Now I hate to be the one to answer my own question, but I feel that in the time it took me to formulate my question in MathJax, I might have arrived at the answer.
First, let's look at why the reduction of degree from two-dimensions to one-dimension for a (joint) sufficient statistic vector for $\theta$ of the Uniform distribution works for symmetrical arguments:
Suppose $X_1,X_2,...,X_n$ is a random sample from the symmetric Uniform distribution $Unif(-\theta,\theta)$. By the factorization theorem, it is easy to verify that the vector $\mathbf Y = (Y_1,Y_2)$ where $Y_1 = X_\left(1\right)$ and $Y_2=X_\left(n\right)$ is a joint sufficient vector of degree two for $\theta$, with $$K_1(Y_1,Y_2;\theta)=(\frac{1}{2})^n \cdot \mathbf 1_{(-\theta,\theta)}(Y_1) \cdot \mathbf 1_{(-\theta,\theta)}(Y_n)$$
From the two indicator functions and from the definition of order statistics, we have that $$-\theta<Y_1<Y_n<\theta \implies \theta>-Y_1 \land \theta>Y_n$$
This allows us to use the maximum function concurrently on $-Y_1$ and $Y_n$ to put a restriction on $\theta$, meaning that this result, $Y^* = max\{-Y_1,Y_n\}$, is such that $$\mathbf 1_{(-\theta,\theta)}(Y_1) \cdot \mathbf 1_{(-\theta,\theta)}(Y_n) = \mathbf 1_{(-\theta,\theta)}(Y^*)$$ is a valid equality.
On the other hand, suppose $X_1,X_2,...,X_n$ is a random sample from the Uniform distribution $Unif(\theta-1,\theta+1)$. By the factorization theorem, it is easy to verify that the vector $\mathbf Y = (Y_1,Y_2)$ where $Y_1 = X_\left(1\right)$ and $Y_2=X_\left(n\right)$ is a joint sufficient vector of degree two for $\theta$, with $$K_1(Y_1,Y_2;\theta)=(\frac{1}{2})^n \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_1) \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_n)$$
From the two indicator functions and from the definition of order statistics, we have that $$\theta-1<Y_1<Y_n<\theta+1 \implies Y_1+1>\theta \land Y_n-1<\theta$$
Because now we have $\theta$ sandwiched between two restrictions ("variables", for our purposes) and without the benefit of appealing to the symmetry of the situation, we have no tools available to ourselves to condense the information provided from $Y_1$ and $Y_2$ any further. Thus, we must concede that the joint sufficient statistics $Y_1$ and $Y_n$ are joint minimal sufficient statistics for $\theta$ for a non-symmetric Uniform distribution. On the other hand, we have also shown that $Y^*=max\{Y_1,Y_n\}$ is the single-dimensional and (thus) minimal sufficient sufficient statistic for $\theta$ for a symmetric Uniform distribution.
Best Answer
By the factorization criterion $$ \mathcal{L}(\theta)=\frac{1}{(2\pi)^{n/2}}\exp\{-\sum_{i=1}^nX_i^2/2 +\bar{X}_n \theta -n\theta^2/2\} $$ $$ \qquad = \exp\{\bar{X}_n\theta-n\theta^2/2\}\times(2\pi)^{-n/2}\exp\{-\sum X_i^2/2\}. $$ So $\bar{X}_n$ is sufficient statistic.