Theorem states that if samples derived from a population with normal distribution has a mean $\mu$ and standard deviation $\sigma$ Then $$Y'= \frac{1}{n}\sum Y_i$$
has mean $\mu$ where $Y_i$'s are random variables for each sample $i=(1,2,...,n)$
Okay, Are we assuming that $Y_i's$ have the same mean here? if $Y_i's$ have different means that what would be according mean of $Y'$
In general, if $Y_i$~$ N(\mu_i,\sigma_i^2)$ then $\displaystyle{E(Y')=E(\frac{\sum_iY_i}{n})=\frac{\sum_i\mu_i}{n}}$
The situation you describe is that all the $Y_i$s are sampled from the same population $N(\mu,\sigma^2)$. So $\displaystyle{E(Y')=E(\frac{\sum_iY_i}{n})=\frac{\sum_i\mu}{n}=\mu}$. (We did not need independence here).
The variance part requires independence of the $Y_i$s. So assuming they are i.i.d (independent, identically distributed), we get $\displaystyle{Var(Y')=Var(\frac{\sum_iY_i}{n})=\frac{\sum_iVar(Y_i)}{n^2}=\frac{\sum_i\sigma^2}{n^2}=\frac{\sigma^2}{n}}$