Is $\operatorname{Aut}(\mathbb{I})$ isomorphic to $\operatorname{Aut}(\mathbb{I}^2)$?

Is $\def\Aut{\operatorname{Aut}}\Aut(\mathbb{I})$ isomorphic to $\Aut(\mathbb{I}^2)$ ? ($\mathbb{I},\mathbb{I}^2$ have their usual meaning as objects in $\mathsf{Top}$).


I show some of one of my attempts. I was trying to show that the two are not isomorphic by considering the elements of order $2$ of both groups. If $\Aut(\mathbb{I})$ has a finite number of elements of order $2$, then $\Aut(\mathbb{I}^2)$ would have more elements of order $2$ (because for each $f\in \Aut(I)$ of order $2$, the functions $f_1,f_2$ given by $f_1(x,y)=(f(x),f(y)),f_2(x,y)=(x,f(y))$ are elements of order $2$ of $\Aut(\mathbb{I}^2)$ ).

Shortly, I found that $\Aut(\mathbb{I})$ has infinitely many elements of order $2$, thus this does not work.

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If you knew that there were only a non-zero finite number of elements of $\text{Aut}(\Bbb I)$ of order $2$, then your approach would do the trick. However, since you know that each map $x\mapsto(1-x^n)^{1/n}$ with $n\in\Bbb Z_+$ is an element of $\text{Aut}(\Bbb I)$ of order $2$, then there are infinitely many such elements in both automorphism groups. Without something more–like showing that the respective infinite collections of order $2$ elements are of different cardinality–this won’t be enough.

Consider instead the function $g:\Bbb I^2\to\Bbb I^2$ given by $$g(x,y)=(1-y,x).$$ You should be able to show that $g$ is an automorphism of order $4$. Does $\text{Aut}(\Bbb I)$ have any elements of order $4$?

Hints: Each auto-homeomorphism on $\Bbb I$ will either be strictly increasing or strictly decreasing. The only strictly increasing auto-homeomorphism on $\Bbb I$ of finite order is the identity function. Consider the multiplicative subgroup $C_2:=\{1,-1\}$ of $\Bbb C^\times,$ and the function $\phi:\text{Aut}(\Bbb I)\to C_2$ taking strictly increasing functions to $1$ and strictly decreasing functions to $-1$. You should be able to show that $\phi:\text{Aut}(\Bbb I)\to C_2$ is a surjective homomorphism. From this, we see that if $f$ is a strictly decreasing auto-homeomorphism of $\Bbb I$ of finite order, then $f$ has even order. How is the order of $f$ related to the order of $f^2$? Is $f^2$ strictly increasing or strictly decreasing? What can we then conclude about the order of $f$?