Let $Y_i \sim N(\mu_i,1)$ and $g(\mu_i) = \frac{\sum_i \mu_i^2}{n}$. $Y_{i's}$ are independent. Find the UMVUE from $g(\mu_i)$.
I know that the distribution belongs to the exponential family and $ \sum_i x_i $ is a sufficient and complete statistic for $\mu_i$
\begin{align*} p(x,\mu_i,1) &= \prod_{i=1}^n \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(x_i - \mu_i)^2\right) \\ &= (2\pi)^{-\frac{n}{2}}\exp\left(-\frac{\sum_i x_i^2}{2} +\sum_i\mu_i \sum_i x_i -\frac{\sum_i \mu_i}{2} \right)\\ &= (2\pi)^{-\frac{n}{2}}\exp\left(-\frac{\sum_i x_i^2}{2}\right)\exp\left(\sum_i\mu_i \sum_i x_i -\frac{\sum_i \mu_i^2}{2}\right) \end{align*}
but I'm stuck in realizing which UMVUE of $g (\mu_i)$, I was thinking of turning $ \mu_i $ into a vector or something, could someone help?
Your conclusion that $\sum\limits_{i=1}^n X_i$ is complete sufficient for $\mu_i$ is incorrect.
Define a random vector $X=(X_1,\ldots,X_n)^T$ where $X_i$'s are independent $N(\mu_i,1)$ variables. Also let $\mu=(\mu_1,\ldots,\mu_n)^T$. Then density of $X$ at $x=(x_1,\ldots,x_n)\in \mathbb R^n$ is
\begin{align} f_{\mu}(x)&\propto \exp\left\{-\frac12\sum_{i=1}^n (x_i-\mu_i)^2\right\} \\&=\exp\left\{-\frac12\sum_{i=1}^n x_i^2-\frac12\sum_{i=1}^n \mu_i^2+\sum_{i=1}^n x_i\mu_i\right\}\quad,\,\mu\in\mathbb R^n \end{align}
Since $\sum\limits_{i=1}^n x_i\mu_i=x^T\mu$ and $f_{\mu}$ is a member of a regular exponential family, it follows that $X$ is complete sufficient for $\mu$. So you only have to find an unbiased estimator of $\frac1n\sum\limits_{i=1}^n \mu_i^2$ based on $X$.
A reasonable place to start is from $E\left[\sum\limits_{i=1}^n X_i^2\right]$.