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 $\phi$?
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.