If $T^n$ is $q$-contractive, $T$ exactly has one fixed point

The attempt to show $T$ is $q$-contractive is doomed, as we will show by an example below.

However any fixed point of $T$ is also a fixed point of $T^n$, and there is only one of these. So if we can show $T$ has a fixed point, we are done.

Let $x \in X$ be the unique fixed point of $T^n$, and consider $T^n(T(x))=T(T^n(x))=T(x)$. But now $T(x)$ is a fixed point of $T^n$, so $T(x) = x$ and $x$ is also a fixed point of $T$.

To show $T$ itself need not be $q$-contractive, consider $T:X\rightarrow X$ on $X=[-1,1]$ defined by $X(x) = |x|$ if $x \lt 0$ and $X(x) = x/2$ if $x \ge 0$. Then $T^2$ is $\frac{1}{2}$-contractive, but $T$ is not $q$-contractive for any $0 \le q \lt 1$.