Does the conjugate of a fixed point of an operator have to be fixed as well?

Let $T$ be an operator with a common fixed point $p$. Does it follow that if $T^n(p) = p$ for some $n \in \mathbb{N}$, then $T^{n’}(p) = p$ for every $n’ \in \mathbb{N}$?

A:

No. You can take
$$T(x) = \frac{x}{1 + x}$$
You see
$$T^2(x) = \frac{x}{(1 + x)(1 + x)}$$
and the denominator goes back to $1$ after taking the $2$-nd root.

A:

Of course not.
Consider $T(x)=x^2$ has a fixed point $T(x)=x$.
Take $n$ to be any number not equal to $1$.
Then, $T^n(x)=x^{2^n}$ is a fixed point of $T$, but $T^{n+1}(x)=x^{2^{n+1}}$ is not since $2^{n+1} eq 1$.

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A fact confirmed in a study undertaken by Dr Ian
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Dr Browning’s research focused on the activity of
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