I think this is a counter-example to what you are asking. Choose a smooth function $a : R \to R$ with $a(0) = 0$, $a(-x) = a(x)$, $a(x) = x^2$ near $x = 0$, $a(x)$ monotonically increasing for $x > 0$, and $a(x)$ everywhere less than $1$. Then if we let $f(x) := (1 - a(x)) x$ we have $f'(0) = 1$, but $f^n(x)$ converges to $0$ for all $x$. This is because $f^n(x)$ is monotonic and bounded, and so approaches a limit, which must be a fixed point and so can only be $0$.
I think this is a counter-example to what you are asking. Choose a smooth function $a : R \to R$ with $a(0) = 0$, $a(-x) = a(x)$, a(x) $a(x) = x^2 x^2$ near $x = 0$, $a(x)$ monotonically increasing for $x > 0$, and $a(x)$ everywhere less than $1$. Then if we let $f(x) := (1 - a(x)) x$ we have $f'(0) = 1$, but $f^n(x)$ converges to $0$ for all $x$. This is because $f^n(x)$ is monotonic, and so approaches a limit, which must be a fixed point and so can only be $0$.
I think this is a counter-example to what you are asking. Choose a smooth function $a : R \to R$ with $a(0) = 0$, $a(-x) = a(x)$, a(x) = x^2 near $x = 0$, $a(x)$ monotonically increasing for $x > 0$, and $a(x)$ everywhere less than $1$. Then if we let $f(x) := (1 - a(x)) x$ we have $f'(0) = 1$, but $f^n(x)$ converges to $0$ for all $x$. This is because $f^n(x)$ is monotonic, and so approaches a limit, which must be a fixed point and so can only be $0$.