Consider a circle $ \mathbb{T} $ and an irrational rotation $ f_{\theta} : \mathbb{T} \to \mathbb{T} \; x \mapsto x+\theta $.
If we define $ f^0_\theta(x)=x $ and $ f^{n+1}_\theta = f^n_\theta \circ f_\theta$ we then can create a set $ F = \{ f^n_\theta : n \in \mathbb{N} \}$ and the set $ F' = \{ f_\theta \circ x : x \in F \} $.
Does $ F = F' $ ?
No. Assuming your $\mathbb N$ includes $0$, $f^0_\theta$ is in $F$ but not in $F'$.