Consider a set of smooth ODE: $$(1)~~~~~~~~~\dot{x} = f(x)$$ with $x \in \mathbb{R}^n$ and $f : \mathbb{R}^n \to \mathbb{R}^n$. Consider also a linear transformaton $\gamma : \mathbb{R}^n \to \mathbb{R}^n$. We say that $\gamma$ is a symmetry if $\gamma (x(t))$ is a solution of $(1)$ when $x(t)$ is a solution, too.
Suppose that $x(t)$ is a solution of $(1)$ and it is $T$-periodic. Consider also that there exists a symmetry $\gamma$ as above.
Can I say that
$$\gamma (x(t)) = x(t + \tau)$$ ?
If so, what about $\tau$?.
Errate Corrige
Thanks to cnick, I realized that I wrote wrongly my question. Here is the correction:
Suppose that $x(t)$ is a solution of $(1)$ and it is $T$-periodic. Can I say that there always exists a symmetry $\gamma$ such that:
$$\gamma (x(t)) = x(t + \tau)$$ ?
If so, what about $\tau$?.
No, your statement will not follow. Consider the ODE: $$ x^{\prime} = -y\\ y^{\prime} = x $$ The solution will be a family of circles around the origin.
By the definitions provided, $\gamma(x,y) = 2(x,y)$ will be a symmetry. But since $\gamma(x(t_1)) \neq x(t_2)$ for any values of $t_1$ or $t_2$ since $\gamma$ is a dilation of the original solution.