Given the following differential equations:

\begin{equation} \begin{aligned} \dot{x}_1 &= f_1(x_1,\ldots,x_n) \\ \vdots \\ \dot{x}_n &= f_n(x_1,\ldots,x_n) \end{aligned} \end{equation}

In a compact way: $$\dot{\hat{x}} = \hat{F}(\hat{x})$$

Let the group $\Psi\subset S_n$, where $S_n$ is a symmetric group.

Suppose the following key property holds: $$\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x}), \ \ \ \ \ \forall \sigma\in\Psi$$ where $P_\sigma$ is the permutation matrix corresponding to $\sigma\in \Psi$.

For example: \begin{equation} \begin{aligned} \dot{x}_1 &= x_1(x_1-1)(x_1+1) +x_2 \\ \dot{x}_2 &= x_2(x_2-1)(x_2+1) +x_1 + x_3 \\ \dot{x}_3 &= x_3(x_3-1)(x_3+1) +x_2\end{aligned} \end{equation}

In this case, we can choose $\sigma = (13)\in \Psi = \{(),(13)\}$, and the corresponding permutation matrix: $$P_\sigma = \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ It is easy to see that $\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x})$

My question:

If $\hat{x}^*$ is a

stableequilibrium point for $\dot{\hat{x}} = \hat{F}(\hat{x})$, can I say $P_\sigma \hat{x}^*$ is also astableequilibrium point, for all $\sigma\in \Psi$?

Note: Easily to see that $P_\sigma \hat{x}^*$ is also an equilibrium point.