Here's a method for computing the action on the integral homology of the associated complex curve $x^{100} + y^{100} = z^{100}$; from this you can derive computations with other coefficients as suggested above.

The composite $[x:y:z] \mapsto [x^{100}:y^{100}:z^{100}] \mapsto [x^{100}:z^{100}]$ exhibits your curve as a cover of $\mathbb P^1$ ramified only over $0$, $1$, and $\infty$. Such covers are classified by the fundamental group of $\mathbb P^1 \setminus \{0,1,\infty\}$, and such (finite) covers are in bijective correspondence with finite sets with a right action of the free group on two generators $F$ (an application of the Riemann existence theorem). I'll write $F = \langle a,b,c | abc = 1 \rangle$ where $a$ is a curve giving monodromy around $0$, $b$ around $1$, and $c$ around $\infty$. Your particular cover comes from $H \backslash F$ where $H$ is the kernel of the homomorphism $F \to \mathbb{Z}/100 \times \mathbb{Z}/100$ sending $a$ to $(1,0)$ and $b$ to $(0,1)$, and the group you're interesting in acting by is a group of deck transformations via monodromy around $\infty$ (so by $(-1,-1)$).

So let's suppose you have a curve classified by $H \backslash F$ and let's compute its homology in an $NH/H$-equivariant fashion.

Take the preimages of $0,1,\infty$; of the real intervals $(-\infty,0),(1,\infty)$; of the open set which is left, which is homeomorphic to an open disc. Taking preimages of these gives you a cell structure on your curve, with three $F$-orbits of points, two $F$-orbits of edges, and one $F$-orbit of 2-cells. Some computation with how edges are glued together allows you to describe the cell structure on your curve as follows:

Points $F \cdot p$, $F \cdot q$, $F \cdot r$ where $p, q, r$ have stabilizers generated by $a$, $b$, and $c$ respectively

Edges $F \cdot \ell$, $F \cdot \ell'$ where $\ell$ is an edge from $p$ to $br$, $m$ is an edge from $q$ to $r$

Two-cells $F \cdot u$, where the boundary of $u$ is attached by the path $m ({}^a \ell)^{-1} \ell ({}^b m)^{-1}$

So the homology is computed $NH/H$-equivariantly by a chain complex
$$
\begin{align*}
\mathbb Z [H \backslash F] \cdot u &\to
\mathbb Z [H \backslash F] \cdot \ell \times
\mathbb Z [H \backslash F] \cdot m\\ &\to
\mathbb Z [H \backslash F / \langle a \rangle] \cdot p \times
\mathbb Z [H \backslash F / \langle b \rangle] \cdot q \times
\mathbb Z [H \backslash F / \langle c \rangle] \cdot r
\end{align*}
$$
Here the boundary of $u$ is $(1-b)m + (1-a)\ell$, the boundary of $\ell$ is $br-p$, and the boundary of $m$ is $r-q$.

If you want cohomology, take Hom out.

This leaves the difficult - but mechanical - process of computing the homology groups with the action of the specific generator that you've listed.