find an unbiased estimator of $\sigma$ based on $\sum X_i^2$ Let $X_1,...X_n$ be iid drawn from $N(\sigma,\sigma^2)$ where $\sigma\gt0$ is the unknown parameter. Find an unbiased estimator of $\sigma$ based on $\sum X_i$, find an unbiased estimator of $\sigma$ based on $\sum X_i^2$, and use these results to determine the minimal sufficient statistic $T=(\sum X_i, \sum X_i^2)$ is complete?
It's easy to find an unbiased estimator based on $\sum X_i$. $E\left(\frac {\sum X_i} n\right)=\sigma$, so $\bar X$ is an unbiased estimator of $\sigma$. But $E(X_i^2)=var(X_i)+(E(X_i))^2=2\sigma^2$, so it's hard to get $E(\sum X_i^2)$ to be $\sigma$. $E(\frac {\sum X_i^2)} {2n})=\sigma^2$, but how can you spit out a $\sigma$?
Neither $\sqrt {\frac {\sum X_i^2)} {2n}}$ nor $\frac n 2\cdot \frac {\sum X_i^2}{\sum X_i}$ seem to work, because you can't put the expectation into the square root, and I don't think you can split an expectation across a numerator and denominator. Or can you? Am I even allowed to include $\sum X_i$ here?
Help

Denote $Y_n = \frac{1}{2n} \sum_{i=1}^n X_i^2$ with $X_i$ iid and follows the normal distribution $N(\sigma,\sigma^2)$, We will contruct an unbiased estimator with the form $\theta(Y_n,n) = \sqrt{Y_n}g(n)$
The estimator $\theta(Y_n,n)$ must satisfy the 2 following conditions for all $n$
$$E(\theta(Y_n,n))= \sigma \tag{1}$$
and
$$\theta(Y_n,n) \xrightarrow{n \rightarrow +\infty} \sigma \tag{2} $$
For the second condition (2), we know already that $\sqrt{Y_n}\rightarrow \sigma$, so it suffices to construct $g(n)$ such that $g(n)\xrightarrow{n \rightarrow +\infty} 1 \tag{3}$
For the first condition (1), we have $Y_n = \frac{\sigma^2}{2n} \sum_{i=1}^n (\frac{X_i}{\sigma})^2 =\frac{\sigma^2}{2n} Z_n$ with $Z_n$ follows the noncentral chi-squared distribution $\chi^2_{n,n}$ of $n$ degrees of freedom and non-centrality parameter $\lambda = n$. The condition (I) is so equivalent to
$$E(\sigma\frac{1}{\sqrt{2n}}\sqrt{Z_n}g(n))= \sigma$$
or
$$g(n)=\frac{\sqrt{2n}}{E(\sqrt{Z_n})}=\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})} \tag{4}$$
For information, $E(\sqrt{\chi^2_{n,n}})$ has a semi-analytical form $$E(\sqrt{\chi^2_{n,n}})=\sqrt{2}\Gamma(\frac{1+n}{2}) \text{Hypergeometric1F1Regularized}(-\frac{1}{2},\frac{n}{2}, -\frac{n}{2}) \tag{5}$$
Now, it suffices to check whether $g(n)$ in (4) satisfies the condition (3). I did check with mathematica and effectively, we have
$$\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})}\xrightarrow{n \rightarrow +\infty} 1 $$
(for information, you can also check this by using Wolfram Alpha by typing the formula (5), for example with $n=1000$ here).
Conclusion: One of the unbiased estimators of $\sigma$ from $\sum_{i=1}^n X_i^2$ is
$$\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})} \sqrt{\frac{1}{2n}\sum_{i=1}^n X_i^2} $$
