An NP-hard problem reduces to its complement?
First note that $\overline X$ is $\operatorname{co\mathcal{NP}}$-hard since $X$ is $\mathcal{NP}$-hard (try to see why).
Since 3SAT is in $\mathcal{NP}$, any problem that can be reduced to 3SAT is in NP as well. So the statement suggests that the $\operatorname{co\mathcal{NP}}$-hard language $\overline X$ is in $\mathcal{NP}$.

On the other hand, saying that $\overline{X}$ is in $\mathcal{NP}$ is equivalent to saying that $X$ is in $\operatorname{co\mathcal{NP}}$. So in total we have $X \in \operatorname{co\mathcal{NP}}$ and $\overline X$ is $\operatorname{co\mathcal{NP}}$-hard. So we get $X \leq_m^p \overline X$ by the definition of hardness.



Here is a bit of intuition of the fact that, if $X$ is $\mathcal{NP}$-hard, then $\overline{X}$ is $\operatorname{co\mathcal{NP}}$-hard.

Let $A$ be an arbitrary language in $\operatorname{co\mathcal{NP}}$. We have to show that $A \leq_m^p \overline{X}$. Let $B := \overline{A}$. Then $B$ is in $\mathcal{NP}$ and hence, $B \leq_m^p X$. This means there is a function $f$ computable in polynomial time, such that for an arbitrary word $x$, $x \in B$ if and only if $f(x) \in X$. 

Now we show that $A \leq_m^p \overline{X}$. For a given word $x$. Using the same reduction $f$, we have $x \in A$ if and only if $x \notin B$ if and only if $f(x) \notin \overline{X}$ if and only if $f(x) \in X$.



Another note. Your intuition was right, but in the sentence


  I think this means $\mathcal{NP}$-hard = $\mathcal{NP}$, and thus $X$ will reduce to its complement in this world.


You should have probably said $\mathcal{NP}$ = $\operatorname{co\mathcal{NP}}$.