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 **stable** equilibrium point for $\dot{\hat{x}} = \hat{F}(\hat{x})$, can I say $P\_\sigma \hat{x}^\*$ is also a **stable** equilibrium point, for all $\sigma\in \Psi$?
> 
> 
> 


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



Does stability of equilibrium point preserved by permutation matrix (symmetry)?
I think that the question is **more suitable for the [Mathematics StackExchange](https://math.stackexchange.com)**, but I'll try to (reasonably fully) answer it here.


In order not to be burdened with unnecessary technicalities, let us assume that $F$ is so regular that Picard's theorem holds. Also, let $\lVert \cdot \rVert$ stand for the Euclidean metric.


Let $\varphi \colon (\alpha, \beta) \to \mathbb{R}^n$ be a nonextendible solution of $\dot{x} = F(x)$ (I deleted hats). That is, one has $\dot{\varphi}(t) = F(\varphi(t))$ for all $t \in (\alpha, \beta)$. We claim that, for any $\sigma \in \Psi$, the mapping 
$$
(\alpha, \beta) \ni t \mapsto P\_{\sigma}\varphi(t) \in \mathbb{R}^n
$$ 
is a solution. Indeed, one has
$$
\dot{(P\_{\sigma}\varphi)}(t) = P\_{\sigma} \dot{\varphi}(t) = P\_{\sigma} F(\varphi(t)) = F(P\_{\sigma} \varphi(t)) \qquad \forall \, t \in (\alpha, \beta).
$$
In the theory of continuous-time dynamical systems, a mapping taking (in our setting) the solutions of one system of ODEs onto the solutions of another system of ODEs is called *conjugacy*.


Recall the definition of (Lyapunov) stability: an equilibrium $x^\*$ is stable if for each $\varepsilon > 0$ there exists $\delta > 0$ such that for any $\tilde{x} \in \mathbb{R}^n$, if $\lVert \tilde{x} - x^\* \rVert < \delta$ then the solution $\varphi(\cdot;\tilde{x})$ of the IVP $\dot{x} = F(x)$, $x(0) = \tilde{x}$, is defined at least on $[0, \infty)$ and the inequality
$$
\lVert \varphi(t;\tilde{x}) - x^\* \lVert < \epsilon
$$
holds for all $t \ge 0$.


We have
$$
\lVert P\_{\sigma} x - P\_{\sigma} y \lVert = \lVert x - y \rVert 
$$
for any $\sigma \in \Psi$ and any $x, y \in \mathbb{R}^n$. This should be enough to conclude the proof.


A proof for asymptotic stability goes in an analogous way.


