Let $X$ be a set and ${\rm Sym}(X)$, the symmetric group on $X$. If $x,y\in X$, is there guaranteed to be an $f\in{\rm Sym}(X)$ such that $f(x)=y$?

Question

Let $X$ be a set and $\operatorname{Sym}(X)$ be the symmetric group on $X$. If $x, y \in X$, is there guaranteed to be an $f \in \operatorname{Sym}(X)$ such that $f(x) = y$?

I think the answer is yes, but I'm having difficulty justifying it logically.

Answer 1

Simply take the involution (transposition) $\;(x\,y)\;$, meaning: the bijective map

$$f:X\to X\;,\;\;f(a):=\begin{cases}y,&a=x\\{}\\x,&a=y\\{}\\a,&a\neq x,y\;\end{cases}\;,\;\;\;\;x\neq y$$

If $\;x=y\;$ then the identity map fulfills what you want.

Answer 2

The underlying set of ${\rm Sym}(X)$ is the set of all bijections from $X$ to $X$. Thus, for a given pair $(x,y)\in X\times X$, select a bijection $f$ such that $y=f(x)$; such an $f$ exists because we can construct it: designate some element $\star\in X$ then send $y$ to $\star$, $x$ to $y$, $\star$ to $x$, and the rest of $X$ to themselves. In other words,

$$f(\xi)=\begin{cases} y &: \xi=x,\\ x &: \xi=\star,\\ \star &: \xi=y,\\ \xi &: \text{otherwise}. \end{cases}$$

But be careful with $\star=x$! In that case, take $f$ to be the involution $(xy)$, as I could have done from the get-go; and if $x=y$, the identity map will do.