Let's assume that we are working over $\mathbb{C}$.

First of all, hypersurfaces in $\mathbb{P}^n$ are unobstructed, so their first-order deformations always correspond to small deformations (deformations over a disk).

As a general fact, when you consider a smooth variety $X$ with a finite group $G$ acting $holomorphically$ on it, the invariant subspace $H^1(X, T_X)^G$, it parametrizes those first-order deformations that preserve the holomorphic $G$-action. This essentially comes from the fact that, being the action of $G$ holomorphic, if you take $\sigma \in G$, then $\sigma_*$ commutes with $\bar{\partial}$ and the Green operator $\boldsymbol{G}$, so if $\varphi(t)$ solves the Kuranishi equation

$\varphi(t)=t + \frac{1}{2}\bar{\partial}^* \boldsymbol{G}[\varphi(t), \varphi(t)]$

for $t$, then $\sigma_*\varphi(t)$ solves the Kuranishi equation for $\sigma_*t$, and $\sigma_{*} \varphi(t) = \varphi(\sigma_*t)$.

Example Let us consider a quintic Fermat surface $X \subset \mathbb{P}^3$ of equation

$x^5+y^5+z^5+w^5=0$.

It admits a free action of the cyclic group $\mathbb{Z}_5$ given as follows

$\xi \cdot (x,y,z,w)=(x, \xi y, \xi^2 z, \xi^3 w) $.

The quotient $Y := X/\mathbb{Z}_5$ is a Godeaux surface (i.e. a surface of general type with $p_g=q=0, K^2=1$ ) with fundamental group $\mathbb{Z}_5$. M. Reid proved that, conversely, every Godeaux surface with fundamental group $\mathbb{Z}_5$ arises in this way and that, moreover, the corresponding moduli space is generically smooth of dimension $8$. Then in this case we have

$\dim H^1(X, T_X)=40$

$\dim H^1(X, T_X)^G=H^1(Y, T_Y)=8$,

since the number of moduli of hypersurfaces keeping the free $G$-action equals the number of moduli of the Godeaux surface $Y$ (well, Horikawa showed that the deformations of quintic surfaces are complicated enough, anyway $40$ is the right number).

Actually, one can say more and check that for every irreducible character $\chi$ of $G$ one has

$\dim H^1(X, T_X)^{\chi} = 8$,

but I do not know any easy interpretation of these eigenspaces in terms of the deformations of the quintic.

Let's assume that we are working over $\mathbb{C}$.

First of all, hypersurfaces in $\mathbb{P}^n$ are unobstructed, so their first-order deformations always correspond to small deformations (deformations over a disk).

As a general fact, when you consider a smooth variety $X$ with a finite group $G$ acting $holomorphically$ on it, the invariant subspace $H^1(X, T_X)^G$, it parametrizes those first-order deformations that preserve the holomorphic $G$-action. This essentially comes from the fact that, being the action of $G$ holomorphic, if you take $\sigma \in G$, then $\sigma_*$ commutes with $\bar{\partial}$ and the Green operator $\boldsymbol{G}$, so if $\varphi(t)$ solves the Kuranishi equation

$\varphi(t)=t + \frac{1}{2}\bar{\partial}^* \boldsymbol{G}[\varphi(t), \varphi(t)]$

for $t$, then $\sigma_*\varphi(t)$ solves the Kuranishi equation for $\sigma_*t$, and $\sigma_{*} \varphi(t) = \varphi(\sigma_*t)$.

Example Let us consider a quintic Fermat surface $X \subset \mathbb{P}^3$ of equation

$x^5+y^5+z^5+w^5=0$.

It admits a free action of the cyclic group $\mathbb{Z}_5$ given as follows: if $\xi$ is a primitive $5$-th root of unity, then

$\xi \cdot (x,y,z,w)=(x, \xi y, \xi^2 z, \xi^3 w) $.

The quotient $Y := X/\mathbb{Z}_5$ is a Godeaux surface (i.e. a surface of general type with $p_g=q=0, K^2=1$ ) with fundamental group $\mathbb{Z}_5$. M. Reid proved that, conversely, every Godeaux surface with fundamental group $\mathbb{Z}_5$ arises in this way and that, moreover, the corresponding moduli space is generically smooth of dimension $8$. Then in this case we have

$\dim H^1(X, T_X)=40$

$\dim H^1(X, T_X)^G=H^1(Y, T_Y)=8$,

since the number of moduli of quintics keeping the free $G$-action equals the number of moduli of the Godeaux surface $Y$ (well, Horikawa showed that the deformations of quintic surfaces are complicated enough, anyway $40$ is the right number).

Actually, one can say more and check that for every irreducible character $\chi$ of $G$ one has

$\dim H^1(X, T_X)^{\chi} = 8$,

but I do not know any easy interpretation of these eigenspaces in terms of the deformations of the quintic.

Let's assume that we are working over $\mathbb{C}$.

First of all, hypersurfaces in $\mathbb{P}^n$ are unobstructed, so their first-order deformations always correspond to small deformations (deformations over a disk).

As a general fact, when you consider a variety $X$ with a finite group $G$ acting $holomorphically$ on it, the invariant subspace $H^1(X, T_X)^G$, it parametrizes those deformations that preserve the holomorphic $G$-action. This essentially comes from the fact that, being the action of $G$ holomorphic, if you take $\sigma \in G$, then $\sigma_*$ commutes with $\bar{\partial}$ and the Green operator $\boldsymbol{G}$, so if $\varphi(t)$ solves the Kuranishi equation

$\varphi(t)=t + \frac{1}{2}\bar{\partial}^* \boldsymbol{G}[\varphi(t), \varphi(t)]$

for $t$, then $\sigma_*\varphi(t)$ solves the Kuranishi equation for $\sigma_*t$, and $\sigma_{*} \varphi(t) = \varphi(\sigma_*t)$.

Example Let us consider a quintic Fermat surface $X \subset \mathbb{P}^3$ of equation

$x^5+y^5+z^5+w^5=0$.

It admits a free action of the cyclic group $\mathbb{Z}_5$ given as follows

$\xi \cdot (x,y,z,w)=(x, \xi y, \xi^2 z, \xi^3 w) $.

The quotient $Y := X/\mathbb{Z}_5$ is a Godeaux surface (i.e. a surface of general type with $p_g=q=0, K^2=1$ ) with fundamental group $\mathbb{Z}_5$. M. Reid proved that, conversely, every Godeaux surface with fundamental group $\mathbb{Z}_5$ arises in this way and that, moreover, the corresponding moduli space is generically smooth of dimension $8$. Then in this case we have

$\dim H^1(X, T_X)=40$

$\dim H^1(X, T_X)^G=H^1(Y, T_Y)=8$,

since the number of moduli of hypersurfaces keeping the free $G$-action equals the number of moduli of the Godeaux surface $Y$ (well, the story about the deformations of quintic surfaces in compliceted enough, as shown by Horikawa, anyway $40$ is the corrected number).

Actually, one can say more and check that for every irreducible character $\chi$ of $G$ one has

$\dim H^1(X, T_X)^{\chi} = 8$,

but I do not know any easy interpretation of these eigenspaces in terms of the deformations of the quintic.

Let's assume that we are working over $\mathbb{C}$.

First of all, hypersurfaces in $\mathbb{P}^n$ are unobstructed, so their first-order deformations always correspond to small deformations (deformations over a disk).

As a general fact, when you consider a smooth variety $X$ with a finite group $G$ acting $holomorphically$ on it, the invariant subspace $H^1(X, T_X)^G$, it parametrizes those first-order deformations that preserve the holomorphic $G$-action. This essentially comes from the fact that, being the action of $G$ holomorphic, if you take $\sigma \in G$, then $\sigma_*$ commutes with $\bar{\partial}$ and the Green operator $\boldsymbol{G}$, so if $\varphi(t)$ solves the Kuranishi equation

$\varphi(t)=t + \frac{1}{2}\bar{\partial}^* \boldsymbol{G}[\varphi(t), \varphi(t)]$

for $t$, then $\sigma_*\varphi(t)$ solves the Kuranishi equation for $\sigma_*t$, and $\sigma_{*} \varphi(t) = \varphi(\sigma_*t)$.

Example Let us consider a quintic Fermat surface $X \subset \mathbb{P}^3$ of equation

$x^5+y^5+z^5+w^5=0$.

It admits a free action of the cyclic group $\mathbb{Z}_5$ given as follows

$\xi \cdot (x,y,z,w)=(x, \xi y, \xi^2 z, \xi^3 w) $.

The quotient $Y := X/\mathbb{Z}_5$ is a Godeaux surface (i.e. a surface of general type with $p_g=q=0, K^2=1$ ) with fundamental group $\mathbb{Z}_5$. M. Reid proved that, conversely, every Godeaux surface with fundamental group $\mathbb{Z}_5$ arises in this way and that, moreover, the corresponding moduli space is generically smooth of dimension $8$. Then in this case we have

$\dim H^1(X, T_X)=40$

$\dim H^1(X, T_X)^G=H^1(Y, T_Y)=8$,

since the number of moduli of hypersurfaces keeping the free $G$-action equals the number of moduli of the Godeaux surface $Y$ (well, Horikawa showed that the deformations of quintic surfaces are complicated enough, anyway $40$ is the right number).

Actually, one can say more and check that for every irreducible character $\chi$ of $G$ one has

$\dim H^1(X, T_X)^{\chi} = 8$,

but I do not know any easy interpretation of these eigenspaces in terms of the deformations of the quintic.