symmetry of a solution of an ordinary differential equation

Let $\gamma$ be a symmetry of the ordinary differential equation (http://planetmath.org/SymmetryOfAnOrdinaryDifferentialEquation) and $x_0$ be a steady state solution of $\dot{x}=f(x)$. If

$$\gamma x_0=x_0$$

then $\gamma$ is called a symmetry of the solution of $x_{\mathrm{0}}$.

Let $\gamma$ be a symmetry of the ordinary differential equation and $x_0(t)$ be a periodic solution of $\dot{x}=f(x)$. If

$$\gamma x_0(t-t_0)=x_0(t)$$

for a certain $t_0$ then $(\gamma ,t_0)$ is called a symmetry of the periodic solution of $x_{\mathrm{0}}\mathit{}\mathrm{(}t\mathrm{)}$.

lemma: If $\gamma$ is a symmetry of the ordinary differential equation and let $x_0(t)$ be a solution(either steady state or periodic) of $\dot{x}=f(x)$. Then $\gamma x_0(t)$ is a solution of $\dot{x}=f(x)$.
proof: If $x_0(t)$ is a solution of $\frac{dx}{dt}=f(x)$ implies $\frac{d{x}_0(t)}{dt}=f(x_0(t))$. Let’s now verify that $\gamma x_0(t)$ is a solution, with a substitution into $\frac{dx}{dt}=f(x)$. The left hand side of the equation becomes $\frac{d\gamma x_0(t)}{dt}=\gamma \frac{d{x}_0(t)}{dt}$ and the right hand side of the equation becomes $f(\gamma x_0(t))=\gamma f(x_0(t))$ since $\gamma$ is a symmetry of the differential equation. Therefore we have that the left hand side equals the right hand side since $\frac{d{x}_0(t)}{dt}=f(x_0(t))$.qed