Fixed points under a finite group action on projective variety Let us have an algebraic action by a finite group G on a complex projective variety $X=\bigcup\limits_{i=1}^N X_i$, whose irreducible components $X_i$ are all smooth and of the same dimension $d$, and whose singular cohomology $H^*(X)$ is generated by algebraic cycles (hence supported in even degrees). Note that variety X itself is NOT smooth (unless N=1).
Is there a fixed-points type theorem that would give us that
$$\text{rank}(H^*(X^G))=\text{rank}(H^*(X))?$$
NB I have EDITED the question. 
 A: Setup
Let $G$ be a finite group acting on a smooth projective
variety $X$, and let
$$
\rho: G \times X \to X
$$
be the action morphism. For any $g \in G$ let $\rho_g$ denote the composition 
$$
X \simeq \{g\} \times X \subset G \times X \xrightarrow{\rho} X
$$
The fixed point locus $X^G \subset X$ is a closed subscheme
(possibly empty):
Edit incorporating Peter McNamara's comment: $X^G$ is always closed when $G$ is an affine algebraic group (link in comments).
When $G$ is finite there's an elementary argument: note that for any $g \in G$, the $g$-fixed points $X^g$ fit
into the cartesian diagram
$\require{AMScd}$
\begin{CD}
X^g @>>> X \\
@VVV @V \mathrm{id} \times \rho_g VV \\
X @> \Delta >> X \times X
\end{CD}
Since $\Delta$ (and $\mathrm{id} \times \rho_{g}$ for that matter) are
closed immersions and closed immersions are compatible with base
change, $X^{g} \to X$ is a closed immersion. Since $X^{G} = \bigcap_{g
\in G} X^{g}$, $X^{G}$ is also a closed subscheme of $X$. 
Remark: This heavily relies on the fact that $G$ is a finite (or
at least discrete) group, which is fine since that was the question,
but I'd be interested in whether/when/how-to-show $X^{G}$ is closed in
the case where $G$ is, say, a positive dimensional linearly reductive
affine group scheme acting on $X$.  
where $\Delta$ is the diagonal. As Sándor pointed out in a
comment it definitely can happen that $X^{G}$ is empty, e.g. if $X$ is
an abelian variety and $G \subset X$ is a finite subgroup acting by
translations.
The fixed point locus is smooth
This is more difficult to prove -- an analytic argument due to Cartan
appears in Algebraic geometry and
topology,
and an alternative approach is Luna's étale slice
theorem.
The idea behind both approaches is to get local enough that the local
geometry at a fixed point $x \in X^G$ is modeled by the tangent space
$T_{X,x}$ with it's linear $G$--action (if $\rho_{g}: X \to X$ is
the action of $g \in G$ on $X$, then $g$ acts on $T_{X,x}$ via $d
\rho_{g}: T_{X, x} \to T_{X, \rho_{g}(x)} = T_{X, x}$). One then
proves that $T_{X^{G}, x} = T_{X,x}^{G}$, the $G$-invariant
subspace.
I won't attempt to go into further detail.
The fixed locus can have arbitrary dimension:
Let $\rho: G \times V \to V$ be a linear representation of a finite group $G$. Then
$\mathbb{P}(V)$ is a smooth projective variety with an induced
$G$-action
$$
\bar{\rho}: G \times \mathbb{P}(V) \to \mathbb{P}(V), \, \, \text{
  where  } \bar{\rho}[v] = [\rho(v)]
$$
Observe that for a non-0 vector $v \in V$, $[v] \in \mathbb{P}(V)^{G}$
if and only if for each $g \in G$ there's a scalar $\lambda_{g} \in
\mathbb{C}^{\times}$ so that $\rho_{g}(v) = \lambda_{g} v$. Evidently
this occurs if and only if $v$ lies in the isotypical summand of a
character (1-dimensional representation) $L$ of $V$.
This observation together with some representation theory of
finite groups yields examples where $\dim X^{G}$ takes arbitrary
values in $0, \dots, \dim X$.
Criteria for $X^{G}$ to be non-empty
There are various ``fixed point theorems'' which guarantee the
existence of, well, fixed points. See for instance the Lefschetz fixed
point
theorem
and the holomorphic Lefschetz fixed point
theorem. The
later is especially powerful and shows for instance that if
$$
H^{i}(X, \mathscr{O}_{X}) = 0 \,\, \text{  for  } i > 0
$$
(for example, if $X$ is Fano or even just rationally connected) then
$X^{G} \neq \emptyset$.
A: The answer to the edited question is no. Take  a double covering  $\pi :X\rightarrow \mathbb{P}^2$ branched along a smooth quartic curve $C$ (so $X$ is a Del Pezzo surface), and $G=\langle \sigma \rangle$, where $\sigma $ is the involution such that $\pi \circ\sigma =\pi $. Then $\operatorname{rk} H^*(X)=10$, but $\operatorname{rk} H^*(X^G)=\operatorname{rk} H^*(C)=8$.
A: There is actually a simple answer. Pick X to be two $\mathbb{C}P^1$'s intersecting transversally, and a $\mathbb{Z}/2$ action that swaps two spheres. Then $\text{rk } H^*(X)=3$ whereas $\text{rk } H^*(X^G) = 1.$
