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We know that if $T$ is a contraction in a complete metric space $X$, then $T$ has a unique fixed point and it holds $$\lim_{k\to\infty} T^k(z)=x \ \ \ \ \forall z\in X.$$

If for some integer $p>0$ we have that $T^p$ is a contraction on $X$ then we can also deduce that $T$ has a unique fixed point.

Actually suppose that $x$ is the unique fixed point for $T^p$. Then $$T^p(x) = x$$ and so $$T(T^p(x)) = T^p(T(x)) = T(x).$$ We have proved that $T(x)$ is a fixed point for $T^p$. But for the uniqueness of $x$ we have that $$T(x)=x$$ that is what we would like to prove.

Now I am wondering if it holds true that $$\lim_{k\to\infty} T^k(z)=x \ \ \ \ \forall z\in X.$$

I suppose yes, because for the standard contraction mapping theorem I have that $$\forall \varepsilon>0 \ \ \exists m: d(T^{pm}(z),x)<\varepsilon$$ so $$\forall \varepsilon>0 \ \ \exists n=pm: d(T^n(z),x)<\varepsilon.$$

Is it correct?

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Since $T^p(x) = x$, $T^{p+1}(x) = T(x)$, hence $d(T^{p+1}(x), T^p(x)) = d(T(x), x)$. By the contractivity of $T^p$, $$ d(T^{p+1}(x), T^p(x)) \le \varkappa d(T(x), x), $$ where $\varkappa \in [0,1)$ is the contraction coefficient. This is possible only if $d(T(x), x)=0$, that is, $T(x) = x$.

Take $z \in X$. For any $l = 0, 1, \dots, p-1$, we have $$ \lim\limits_{m \to \infty} T^{mp + l}(z) = \lim\limits_{m \to \infty} T^{mp}(T^l(z)) = x, $$ from which it follows that $$ \lim\limits_{n \to \infty} T^{n}(z) = x. $$

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