Is $x^x$ unique over the positive rationals? Is the function $f(x) = x^x$ unique over the positive rationals? That is, do there exists two different positive rational numbers $x$ and $y$ so that $x^x = y^y$ ?
 A: Here's a family of such rational solutions:
$$
x = \left(\frac{m-1}{m}\right)^{m}, \quad y = \left(\frac{m-1}{m}\right)^{m-1}
$$
for integers $m \geq 2$.

How to get these solutions: assume that $x = b/a, y = c/d$ with $\gcd(a, b) = \gcd(c, d)=  1$. Then the equation $x^x = y^y$ gives that the prime factors of $a$ (resp. $b$) should coincides with $c$ (resp. $d$). Also, setting $a = p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}, b = q_{1}^{f_{1}}\cdots q_{s}^{f_{s}}, c = p_{1}^{e_{1}'}\cdots p_{r}^{e_{r}'}, d = q_{1}^{f_{1}'}\cdots q_{s}^{f_{s}'}$ we can prove $e_i' / e_i = f_i' / f_i = bc/ad$ for all $i$, so that $y = x^t$ for some $t\in \mathbb{Q}_{>0}$. From this, we have $x = t^{-(1/(t-1))}$, and $t = (m-1)/m$ makes $x, x^t \in \mathbb{Q}$.
A: I followed the idea of @ProfessorVector. The resulting pairs should be the same as given by @SeewooLee.
Consider the form $y=x^r$ and suppose $r\in\mathbb{Q}$ and $r>1$ without loss of generality.
Claim that if $n:=\frac1{r-1}\in\mathbb{N}$, then
$$x=r^{-n},\,y=r^{-rn}$$
is a pair in need.
At first, by $rx^{r-1}=1$, we know that
$$y^y=x^{rx^r}=x^x.$$
It is clear that $x\in\mathbb{Q}$. For $y$, we note that
$$y=x^r=r^{\frac{r}{1-r}}=\frac1r\left(\frac1r\right)^n\in\mathbb{Q}.$$
This proves the claim. $\square$
From this result, we know that $r=2,\,\frac32,\,\frac43$... could be
possible choices, in which cases $(x,y)=(\frac12,\frac14),\,(\frac49,\frac8{27}),\,
(\frac{27}{64},\frac{81}{256})$...
