Yes, $\omega_0^*$ is indeed the order type of the negative integers. The idea is just to "write it upside down" and get a new order type.
Sometimes, however, when writing the order type in reverse we get the same result, for instance if you take $\{0,1,2,3\}$ the natural order $0<1<2<3$ and its inverse $3>2>1>0$ are of the same order type.
Furthermore, if you take the integers you have the same result, that is $\mathbb Z$ with the usual ordering $<$ and with the inverse ordering $>$ are of the same order type (the function $x\mapsto -x$ is the isomorphism needed to show that).
From the above example, one can infer the result for the rational numbers. The inverse order means that $q_1 then $q_2>q_1$, and as before $x\mapsto -x$ is the needed isomorphism to prove that.
Added: We say that $a if $\langle a,b\rangle\in <$, inverse orders are essentially $\langle b,a\rangle$ when $\langle a,b\rangle\in <$.
Suppose $\langle A,R\rangle$ and $\langle B,S\rangle$ are two ordered sets, an order isomorphism $f$ is a function with the following properties:
- $f\colon A\to B$ is a bijection;
- $aRb\iff f(a)Sf(b)$, that is to say: $\langle a,b\rangle\in R\iff \langle f(a),f(b)\rangle\in S$.
From this follows that an isomorphism preserves properties such as "$a$ is minimal if and only if $f(a)$ is minimal".
This means that $\omega$ and $\omega^*$ are not isomorphic since $0$ is minimal in the non-negative integers, but $f(0)$ is maximal in the non-positive integers.