The point of this answer is to give a short proof of the following result.
Theorem 1: Let $X$ and $Y$ be smooth $n$-dimensional projective varieties over a field $k$ of characteristic zero; let $B$ be a smooth connected variety; and let $\phi: X \times B \to Y$ be a morphism. Then the induced map $H^q(Y, \Omega_Y^p) \to H^q(X \times \{ b \}, \Omega_X^p)$ is a constant function of $b$. (If the base field is not algebraically closed, this requires a little more language to say correctly; I'm going to gloss over this.)
We need the following, very deep theorem:
Theorem 2: If $X$ is a smooth projective variety over a field of characteristic zero, then the map $H^q(X, \Omega^p) \to H^q(X, \Omega^{p+1})$ induced by $d$ is $0$.
This is usually proved by Hodge theory.
As a corollary of Theorem 1, if $\mathrm{Aut}(X)$ is connected, then $\mathrm{Aut}(X)$ acts trivially on $H^q(X, \Omega^p_X)$. As a corollary of the corollary, $\mathrm{PGL}_{n+1}$ acts trivially on $H^n(\mathbb{P}^n, \omega)$.
Proof (Sketch):
We may immediately assume that $B$ is one dimensional, as any two points in a connected variety may be joined by a chain of smooth affine curves. Also, the statement is local on $B$, so we may pass to a smaller base curve whenever it is convenient.
Let $\theta$ be a class in $H^q(Y, \Omega^p_Y)$. Let $\eta = \phi^* \theta \in H^q(X \times B, \Omega^p_{X \times B})$. For any $b \in B$, we can restrict $\eta$ to $X \times \{ b \}$, and get a class in $H^q(X, \Omega^p_X)$. Call this $\eta(b)$. We want to show that $\eta(b)$ is a constant function of $b$. In other words, $\eta(b)$ is a section of $H^q(X, \Omega_X^p) \otimes \mathcal{O}_B$ and we want to show that it is in $H^q(X, \Omega_X^p)$. Since $H^q(X, \Omega_X^p) \otimes \mathcal{O}_B$ is a trivial vector bundle, it makes sense to talk about $d \eta(b)$ (just take $d$ of each component), and we will show that $d \eta(b)=0$.
In case you would like to see this done in more technical language, here it is: We can think of $H^q(Y, \Omega_Y^p)$ as $R^q \pi_* \Omega_Y^p$ where $\pi$ is the map $Y \to \mathrm{Spec} \ k$. So we can pull back to $\theta$ to a class $\eta$ in $R^q \pi_2^* \Omega^p_X$ where $\pi_1$ and $\pi_2$ are the projections of $X \times B$ onto its factors.
We have a canonical map $\Omega^p_{X \times B} \to \Omega^p_{X \times B/B}$ (restricting to the vertical fibers). So we get a class in $R^q (\pi_2)_{\ast} \Omega^p_{X \times B/B}$. Also, $\Omega^p_{X \times B/B} \cong \pi_1^* \Omega_X^p$. Finally, if you think about how pushforwards and pullbacks work in products, then $R^q (\pi_2)_{\ast} \pi_1^{\ast} \Omega^p_X \cong H^q(X, \Omega^p_X) \otimes \mathcal{O}_B$. So that's how we get a class in $H^q(X, \Omega^p_X) \otimes \mathcal{O}_B$.
So, we want to compute $d \eta(b)$.
Passing to a local neighborhood on $B$, we may assume that $\Omega^1_B$ is free, with generator $dz$ for $z \in \mathcal{O}(B)$.
On $X \times B$, we have
$$\Omega^p_{X \times B} \cong \pi_1^* \Omega_X^p \oplus \ \pi_1^* \Omega_X^{p-1} \otimes \pi_2^* \Omega_B^1 \quad (*)$$
($B$ is one dimensional; otherwise there would be more terms.)
Locally, write $\eta = \alpha + \beta dz$ where $\alpha$ and $\beta$ are sections of $\pi_1^* \Omega^p_X$ and $\pi_1^* \Omega^{p-1}_X$. So we get $\alpha$ and $\beta$ in $R^q (\pi_2)_{\ast} \Omega^{p}_{X \times B/B}$ and $R^q (\pi_2)_{\ast} \Omega^{p-1}_{X \times B/B}$. (More precisely, $\eta$ is represented by a Cech cocycle $U_{i_0} \cap U_{i_1} \cap \cdots \cap U_{i_p} \mapsto \eta_{i_0 i_1 \cdots i_p}$. Write each $\eta_{I}$ as $\alpha_I + \beta_I dz$. Then $\alpha_I$ and $\beta_I$ are Cech cocycles for the corresponding cohomology groups.)
Write $d'$ for the derivation $\Omega^p_{X} \to \Omega^{p+1}_{X}$ and $d''$ for the analogous derivation on $B$. Using the isomorphism $(*)$, we have $d=d'+d''$. In these terms, we have
$$d \eta = d'\alpha + d'' \alpha + d'(\beta) dz. \quad (**)$$
By Theorem 2, $d'(alpha)$ and $d'(\beta)$ are $0$. So $d \eta$, which is a class in $R^q (\pi_2)_{\ast} \Omega^{p+1}_{X}$, is actually a class in $R^q (\pi_2)_{\ast} \pi_1^* \Omega^{p}_{X} \otimes \Omega^1_B$ and, as such, is equal to $d'' \alpha$.
Now, $\eta(b)$ is the restriction of $\eta$ to $X \times \{ b \}$.
As $dz$ is $0$ on the vertical fibers, $\eta|_{X \times \{ b \}} = \alpha|_{X \times \{ b \}}$.
We wanted to compute $d \eta(b)$, where the $d$ is with respect to the $B$-variables.
This makes it plausible, and I leave it to you to check, that $d \eta(b)$ is $d'' \alpha$.
So far, this argument has been valid for any $\eta \in R^q (\pi_2)_{\ast} \Omega^p_{X \times B}$.
Now we use that $\eta = \phi^* \theta$. By Theorem 2, $\theta$ is closed. So $d \eta = d \phi^* \theta = \phi^* d \theta = 0$.
Looking at $(**)$, and at the compatibilities we have just checked, we see that $d \eta(b)=0$, as desired.
Comments:
(1) The analogous result in differential geometry is that, if $B$ is connected, and $\phi : X \times B \to Y$ is a smooth map, then the induced maps $H^{k}(Y) \to H^k(X)$ are the same for every point of $B$. The reader might enjoy writing down a deRham proof of this fact, and seeing how it relates to the above argument.
(2) More generally, let $\pi: \mathcal{X} \to B$ be a smooth projective map with $B$ connected; let $\phi: \mathcal{X} \to Y$ be any morphism with $Y$ projective; let $\theta \in H^q(Y, \Omega_Y^p)$; let $\eta = \phi^* \theta \in R^q \pi_{\ast} \Omega^p_{\mathcal{X}}$. Then we want to say that $\eta(b)$ is a constant function of $b$. The right way to say this is that there is a connection on $R^q \pi_{\ast} \Omega^p_{\mathcal{X}}$ which annihilates $\eta$. This is called the Gauss-Manin connection, and I have basically walked you through how to compute it for a trivial family.
(3) Theorem 2 is trivial for $p=n$. However, even if you only care about Theorem 1 for $p=n$, then our argument uses Theorem 2 applied to $\beta$, which is an $n-1$ form. So we still need to use a nontrivial case of Theorem 2, even in that case. (This is what I missed the first time I wrote this up.)
Best Answer
So it seems I got it backwards: If you start with applying Thm 7.6 to $H^i(X, \mathcal{F})$ you get:
$$H^i(X, \mathcal{F}) = Ext^{n-i}(\mathcal{F}, \omega_X^\circ)' = Ext^{n-i}(\mathcal{O}_X, \mathcal{F}^\vee \otimes \omega_X^\circ)' \cong H^{n-i}(\mathcal{F}^\vee \otimes \omega_X^\circ)'$$