Is there an irrational number $a$ such that $a^a$ is rational?
By continuity of the function $x^x$, there is an $a$ such that
$$a^a=2.$$
This number is irrational. Otherwise, let $a$ be the irreducible fraction $p/q$ and
$$\left(\frac pq\right)^{p/q}=2,$$ is equivalent to $$p^p=2^qq^p,$$
which implies that $p$ is even and $q$ is even, a contradiction.
By the way, as
$$\ln(a^a)=a\ln(a)=\ln(a)e^{\ln(a)},$$
we have
$$\color{green}{a=e^{W(\ln(2))}}$$
where $W$ is the Lambert function.
If $a^a=2$, then $a$ is irrational: If $a=p/q$, then $(p/q)^p=2^q$ is an integer, so $p/q$ is an integer.
Consider the unique (positive) solution $a$ to $x^x = 2$. If $a$ were rational, say, $a = \frac{p}{q}$, $p$ and $q$ are positive integers such that $\gcd(p, q) = 1$, we would have $$\left(\frac{p}{q}\right)^{p / q} = 2 ,$$ and rearranging gives $$p^p = q^p 2^q .$$ Since there is no integer $n$ such that $n^n = 2$, we must have $q > 1$ and hence $2 \mid p^p$. Because $2$ is prime, we have $2 \mid p$. So, $2$ occurs an even number of times the prime factorization of $p^p$ and likewise of $q^p$. Since $p^p = q^p 2^q$, we must have $2 \mid q$, but now $2 \mid p, q$, and this contradicts $\gcd(p, q) = 1$. Thus, $a$ is irrational but $a^a$ is rational (in fact, an integer).