A question regarding complete metric spaces

Question

Let $(M,d)$ be a complete metric space. Let $T:M\to M$ such that $T^n$ is a strict contraction(There is a $0\le k\lt 1$ such that $d(T(x),(T(y))\le kd(x,y)$ for every $z$ and $y$ belonging to $M$ ), $n>1$ but $T^{n-1}$ is not a strict contraction. How many fixed points can $T$ have? If possible, provide examples in each case. I've no idea how to approach this problem. Solutions will be really helpful.

Answer 1

One and only one: the same of $T^n$. Let $x\in M$ be the unique fixed point of $T^n.$

Now, we have that $T^n(T(x))=T(T^n(x))=T(x),$ and this implies that $T(x)$ is a fixed point of $T^n,$ so $T(x)=x$ by uniqueness of fixed point. This proves that x is a fixed point of $T.$

On the other hand, if $T(y)=y,$ iterating T we have that also $T^n(y)=y,$ and by uniqueness of the fixed point we conclude that $y=x,$ so it is unique.

Answer 2

If $T^n$ is a contraction there is exactly one fixed point $x^*$.

Suppose $x$ is a fixed point of $T$, then $T^n x = x$ and hence $x=x^*$, so $T$ can have at most one fixed point.

Noting that $T^n(T^k x^*) = T^k x^*$, we see that $T^k x^* = x^*$ for all $k\ge 1$.

If we let $f(x) = (x_2,1)$, we see that $f^2$ is a contraction, but $f$ is not.