If a function has an infinite amount of automorphisms would that imply that it is periodic?

Let $\phi(x)$ be continuous and differentiable everywhere and its automorphisms be bijections $\gamma_{n}$ such that $\phi(\gamma_{n}(x))=\phi(x)$. If the order of $Aut(\phi)$ is infinite, would that mean that $\phi(x)$ is periodic?

I can think of an example of the top of my head: $\phi(x)=sin(x)$ with automorphisms of the form $\gamma_{n}(x)=x+2\pi n$. Could this be applied to any general function $\phi$?

• What is, in this case, an automorphism of $\phi$? A diffeomorphism $\gamma$ of the domain of $\phi$ such that $\phi \circ \gamma = \phi$? – Daniel Fischer Nov 7 '17 at 16:08
• Yes, the pdf that im reading describes it as a bijection $\alpha$ such that $\phi(\alpha x)=\phi(x)$ – aleden Nov 7 '17 at 16:09
• If there are no continuity requirements on $\alpha$, it would suffice that some value is attained by $\phi$ infinitely often. – Daniel Fischer Nov 7 '17 at 16:11
• If we don't have any further condition on $\gamma$, then it's wrong. Let $\phi : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$. Then for any $a \in \mathbb{R}$ the function $\gamma: \mathbb{R} \to \mathbb{R}, \gamma(x) = -x$ for $x \in \{a, -a\}$ and $\gamma(x) = x$ for $x \notin \{a, -a\}$ is such an automorphism. – S. M. Roch Nov 7 '17 at 16:12
• How would $\phi(x)=x^2$ have an infinite amount of automorphisms? – aleden Nov 7 '17 at 16:17

If we don't have any further condition on $\gamma$, then it's wrong. Let $\phi : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$. Then for every $a > 0$ the function $$\gamma: \mathbb{R} \to \mathbb{R},\hspace{3mm} x \mapsto \begin{cases}-a & x \in \{a, -a\} \\ x & x \notin \{a, -a\}\end{cases}$$
is such an automorphism. Hence we have infinitely many automorphisms. But $\phi$ isn't periodic, so this is a counterexample.