Given $X_1,...,X_n \sim Gamma(\alpha,1/\alpha)$ random variables for some $\alpha>0$, let $\hat{\alpha}$ be a consistent estimator of the sample average $\bar{X}_{n}$ of the sample in terms of $\alpha$.
The following consistent estimator holds since $\bar{X}_{n}$ converges to ${\alpha}^2$ in probability as $n\rightarrow\infty$
$$\hat{\alpha}=\sqrt{\bar{X}_{n}}$$
The central limit theorem (CLT) denotes that
$$\frac{\bar{X}_{n}-\mu}{\sigma/\sqrt{n}}\stackrel{a}\sim N(0,1)$$
For $\bar{X}_{n}\sim Gamma(\alpha,1/\alpha)$
$$\mu=\frac{\alpha}{\beta}=\frac{\alpha}{1/\alpha}=a^2$$
and
$$\sigma^2=\frac{\alpha}{\beta^2}=\frac{\alpha}{(1/\alpha)^2}=a^3\Rightarrow\sigma=\sqrt{a^3}$$
Therefore, applying the CLT
$$\frac{\bar{X}_{n}-\alpha^2}{\sqrt{a^3}/\sqrt{n}}\stackrel{a}\sim N(0,1)\Rightarrow\bar{X}_{n}\stackrel{a}\sim N\big(\alpha^2,\frac{\alpha^3}{n}\big)$$
So far so good, but now instead $\hat{\alpha}=\sqrt{\bar{X}_{n}}$. This calls for the delta method
$$\sqrt{n}(g(X_n)-g(\theta))\stackrel{D}\rightarrow N\big(0,\sigma^2(g'(\theta))^2\big)$$ $$\Rightarrow\sqrt{n}(\sqrt{\bar{X}_{n}}-\sqrt{\alpha})\stackrel{D}\rightarrow N\big(0,\frac{\alpha^3}{n}(\sqrt{a})^2\big)$$ $$\Rightarrow\sqrt{n}(\sqrt{\bar{X}_{n}}-\sqrt{\alpha})\stackrel{D}\rightarrow N\big(0,\frac{\alpha^4}{n}\big)$$ $$\Rightarrow\sqrt{n}(\sqrt{\bar{X}_{n}})\stackrel{D}\rightarrow N\big(\sqrt{\alpha},\frac{\alpha^4}{n}\big)$$ $$\Rightarrow\sqrt{n}\hat{\alpha}\stackrel{D}\rightarrow N\big(\sqrt{\alpha},\frac{\alpha^4}{n}\big)$$ $$\Rightarrow\hat{\alpha}\stackrel{D}\rightarrow N\big(\sqrt{\alpha},\frac{\alpha^4}{n^2}\big)$$
Where did I go wrong?
Apparently, $Var(\hat{\alpha})=a^4/n^2$ is wrong since the answer should not include $n$, or at least $n$ should be somehow $1$ and $Var(\hat{\alpha})≠a^4$ either. I feel like it's a simple mistake which I can't find so perhaps someone else can shed some light on it.