Are these mapping tori different? We can identify the mapping class group of the torus $ T^2 $ with $ SL_2(\mathbb{Z}) $. Let $ k $ be an integer. Consider the mapping
$$
\phi_k=
\begin{bmatrix}
1 & k \\
0 & 1
\end{bmatrix}
$$
and the corresponding mapping torus $ N_k $ as well as the mapping
$$
\theta_k=
\begin{bmatrix}
-1 & k \\
0 & -1
\end{bmatrix}
$$
and the corresponding mapping torus $ M_k $.
Is it is the case that $ N_k $ and $ M_k $ are not diffeomorphic?
What are some topological invariants that distinguish these manifolds? Note that different elements of the mapping class group do not always have distinct mapping tori. For example $ N_k $ and $ N_{-k} $ are diffeomorphic (indeed this must be true since the mapping classes are conjugate in $ SL^{\pm}_2(\mathbb{Z}) $ through conjugation by $ diag(1,-1) $). Even non conjugate mapping classes can have diffeomorphic mapping tori, see
https://mathoverflow.net/questions/241822/homeomorphic-but-non-conjugate-mapping-tori
It was noted here
https://math.stackexchange.com/a/3791368/758507
that $ N_{-2k} $ double covers $ M_k $. As an example of this, $ N_0 $ is the torus $ T^3 $, a flat manifold that can be viewed as the unit tangent bundle of $ T^2 $, and it double covers $ M_0 $, a flat manifold that can be viewed as the unit tangent bundle of the Klein bottle.
Is it true that every $ N_k $ is distinct from every $ M_k $ and furthermore that $ N_k \cong N_{k'} $ and $ M_{k} \cong M_{k'} $ if and only if $ k=\pm k' $?
Note that
$$
\pi_1(N_k) \cong \mathbb{Z}^2 \rtimes_{\phi_k} \mathbb{Z}
$$
and
$$
\pi_1(M_k) \cong \mathbb{Z}^2 \rtimes_{\theta_k} \mathbb{Z}
$$
In principal this should be enough to distinguish them but to be honest its not obvious to me how to decide if two semidirect products are isomorphic.
EDIT: Perhaps I should have written in the original post some things I already know:
$ N_k $ is an orientable 3 manifold (for example because it the quotient of the Heisenberg group by a lattice) and is a (principal) circle bundle over the torus $ T^2 $ (which is also orientable). So the Seifert invariants are
$$
(k,(o_1,1))
$$
Note that there are no exceptional points because this Siefert fibration corresponds to an actual fiber bundle
$$
S^1 \to N_k \to T^2
$$
with first Chern class/ Euler class $ k $. $ N_k $ is a three dimensional nilmanifold so the fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $ must have abelianization with free rank (first betti number) $ 2 $. See for example theorem 5.2 of
https://arxiv.org/abs/0903.2926
For an extremely clear explanation of the full computation of
$$
 H_1(N_k) \cong \mathbb{Z}^2 \oplus \mathbb{Z}_k 
$$
given in the comment by Michael Albanese see the answer from Michael Albanese linked above. In general I am very satisfied with the state of my knowledge about $ N_k $. As a result I was already aware of the fact that all $ N_k, N_{k'} $ are distinct being distinguished, for example, by the torsion in their first homology, except for the case $ k=\pm k' $ in which case the mapping classes are conjugate and they are diffeomorphic $ N_k \cong N_{-k} $ as I mentioned in my original post.
So really it is $ M_k $ that I don't understand.
I am aware of certain statements like: The Seifert invariants of $ M_k $ are
$$
\{b, (o_1,0),(2,1),(2,1),(2,1),(2,1) \}
$$
but I do not understand why or how to show this or why distinct Seifert invariants in this case means the manifolds are distinct. I'm not even sure how $ b $ is related to $ k $. Maybe $ b=k-2 $?
(In general the same manifold can have different sets of Seifert invariants even here one of these manifolds ($ k=0 $ which perhaps corresponds to $ b=-2 $?) is the same as the unit tangent bundle over the Klein bottle which is an actual circle bundle and thus has an alternative presentation in terms of Seifert invariants $ \{0, (n_2,2) \} $).
I am very curious to hear more about the computation of $ H_1(M_k) $ suggested by Michael Albanese. And perhaps any other computations that can distinguish the different $ M_k $.
In the meantime I will take the suggestion from Moishe Kohan and read more of Peter Scott's article to see if that clears things up about Seifert invariants for $ M_k $.
 A: Besides the invariants suggested in the comments, for a Nil-manifold $M_\phi$ defined by $\phi \in \text{SL}_2(\mathbb Z)$ such that $\text{trace}(\phi)=\pm 2$, the conjugacy classes of $\phi$ and $\phi^{-1}$ in the group $\text{SL}_2(\mathbb Z)$ do indeed form a topological invariant of $M_\phi$ (unlike the situation in the link you provided). Since the conjugacy classes of $\{\phi_{\pm k}\}$ and of $\{\theta_{\pm k}\}$ are all different, the mapping toruses of $M_k$ and $N_k$ are all in different homeomorphism classes.
Here's a sketch of the proof.
First, in any 3-dimensional Nil-manifold $M$, the torus fibers of all fibrations of $M$ over $S^1$ form just a single isotopy class of embedded toruses in $M$ (this is proved by arguments of 3-manifold topology; this is the part which is quite different from the link you provided).
Next, fixing one torus fiber, and choosing the "direction" transverse to the torus for the fibration over the circle, the monodromy map of the fibration falls into one of an inverse pair of mapping classes of the fiber (this is standard for all fibrations over a circle).
Finally, choosing any homeomorphism from the standard torus $T^2$ to a torus fiber, the resulting conjugacy class in $\text{SL}_2(\mathbb Z)$ falls into one of two inverse conjugacy classes, independent of all choices.
