With $\Delta=\{(x,\cdots,x)\}$, the set $\{f:X^n\to X^n| \ f\ \text{bijective and}\ f(\Delta)=\Delta\}$ is infinite for all $n\ge1$ if $X$ is infinite

Question

Given $X\ne\emptyset, n\in\Bbb Z^+$ let $\Delta=\{(x,\cdots,x)| x\in X\}\subset X^n.$ Define then $\Delta_n(X)=\{f:X^n\to X^n | \ f\ \text{bijective and}\ f(\Delta)=(\Delta)\}.$ Is it true that if $X$ is infinite then $\Delta_n(X)$ is too for all $n\ge1$?

My initial remark: if $\Delta_1(X)$ is infinite then a fortiori so is $\Delta_n(X)$ for $n>1$, and the claim is true; if $\Delta_1(X)$ is finite then the claim is false. Is it a correct observation? From this it follows we may assume $n=1$ without loss of generality.

Let $\{x_i\}_{i\in\Bbb Z^+}\subset X$, and define a series of functions $\{f_i\}_{i\in\Bbb Z^+}$ so that for all $i,k\ge1$ one has $f_i(x_i)=x_i, f_i(x_k)=x_{i+k}$ and $f_i(x)=x$ for $x\not\in\{x_i\}_{i\in\Bbb Z^+}$. Then all the $f_i$ are elements of $\Delta_1(X)$, which is thus infinite. Am I right?

Answer 1

You can always extend each bijective map from $\Delta$ to $\Delta$ to a bijective map from $X^n$ to $X^n$ and if $X$ is infinite than $\Delta$ is infinite and so there is always an infinite sets of different maps to $\Delta$ to $\Delta$. (Infact you can prove that if $A\neq \emptyset$ is infinite than the sets of bijective map from $A$ to $A$ is infinite because $A\times A$ is infinite and you can choose for any $(a,b)\in A\times A$ the bijective function that maps a in $b$ and $b$ in $a$)

In any case your reasoning is correct.