I am confused about the following easy stuff,
Let $$\mathbf{x} =\begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix}^T .$$ Suppose I have the following monomial $$\mathbf{x}^{\alpha}=x^{\alpha_1}_1 x^{\alpha_2}_2\cdots x^{\alpha_n}_n,$$ corresponding to $$\alpha = \begin{bmatrix} \alpha_1 & \cdots & \alpha_n\end{bmatrix}^T$$ Suppose there is a permutation $\gamma\in \Gamma$ ($\Gamma$ a symmetric group) acting on $\mathbf{x}$ such that $$(\gamma \mathbf{x})^{\alpha} = x^{\alpha_1}_{\gamma(1)} x^{\alpha_2}_{\gamma(2)}\cdots x^{\alpha_n}_{\gamma(n)}.$$
If we rearrange the order such that the index is increasing, i.e., $x_1x_2\ldots x_n$, then the new power $\alpha$, $\alpha'$ should be $$\alpha' = M_\gamma \alpha,$$ i.e., the multiplication of the matrix representation of $\gamma$ (permutation matrix corresponding to $\gamma$) and the vector $\alpha$.
For example $\gamma = (13)$ so $\gamma(1) = 3$ and $\gamma(3) = 1$, $$\mathbf{x}^\alpha = x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}$$ $$M_\gamma = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}^T$$ so $$(\gamma\mathbf{x})^\alpha = x_1^{\alpha_3}x_2^{\alpha_2}x_3^{\alpha_3}.$$ In this case $\alpha = \begin{bmatrix} \alpha_1 & \alpha_2 & \alpha_3\end{bmatrix}^T$ and $\alpha' = \begin{bmatrix} \alpha_3 & \alpha_2 & \alpha_1\end{bmatrix}^T$.
Does this result hold for any permutation? for example $(123)$ or general order.
Yes, the result is true in general. Writing everything out explicitly, your notation $\textbf{x}^\alpha$ can be written as $$ \textbf{x}^\alpha:=\prod_{i=1}^n x_i^{\alpha_i}. $$ Furthermore, $\gamma\textbf{x}$ can be written out explicitly as $$ \gamma\textbf{x}:=\prod_{i=1}^n x_{\gamma(i)}. $$ Your question is asking whether $(\gamma \textbf{x})^{\alpha}=\textbf{x}^{\alpha'}$ where $\alpha'=M_\gamma \alpha$ and you are using $$ (M_\gamma)_{ij}=\begin{cases}1,&i=\gamma(j)\\0,&else.\end{cases} $$
Proof that $(\gamma \textbf{x})^{\alpha}=\textbf{x}^{\alpha'}$.
First, observe that $\alpha'$ can be written as follows: $$ \alpha'_i=\sum_{j}M_{ij}\alpha_j=\alpha_{\gamma^{-1}(i)}. $$ Therefore, $$ (\gamma\textbf{x})^\alpha=\prod_{i=1}^n x_{\gamma(i)}^{\alpha_i} =\prod_{j=1}^n x_{j}^{\alpha_{\gamma^{-1}(j)}}=\textbf{x}^{\alpha'}. $$ In the middle equality, we have reindexed the product using $j=\gamma(i)$, so that $i=\gamma^{-1}(j)$. This reindexing is valid since $\gamma$ is a permutation and the variables $x_i$ commute with one another, so the order the product is written in does not matter.