I'm looking to validate my solution or understand why it's incorrect.
Attempt: By writing the pdf of the distribution in the form $$f(x;\theta)=\exp\left(x\log\left(\frac{\theta}{1-\theta}\right) + 2\log(1-\theta) + \log{2\choose{x}}\right)$$
We see that the distribution belongs to exponential family so $\sum_{i=1}^nX_i$ is CSS for $\theta$.
Now by additive property of Binomial distribution, $$Y=\sum_{i=1}^nX_i$$
has $Binomial(2n,\theta)$ distribution.
Thus $$E[Y^2]=Var(Y)+E[Y]^2 = 2\theta -2\theta^2 + 4n^2\theta^2$$
and $$E[\tfrac{Y}{n}]=2\theta$$
so after a little algebra we get that
$$E\left[h(Y)=\frac{Y^2 - \tfrac{Y}{n}}{4n^2-2}\right]=\theta^2$$
Lehmann-Schaeffe tells us that $$E[h(Y)|Y]=h(Y)$$ is thus UMVUE of $\theta^2$. Is it correct?