Question. Suppose $X$ is a smooth irreducible projective variety over a field $k$. Let $\omega_{X/k}=\bigwedge^{\dim X} \Omega_{X/k}$ be the canonical sheaf with associated divisor $K_X$. Suppose there is an integer $m$ so that $mK_X$ is very ample. Why is there an injection from $\operatorname{Aut}(X/k)$ to the automorphisms of the global sections of $mK_X$?
My understanding. I think I can prove that if $f$ is a $k$-automorphism of $X$, then $f^*\Omega_{X/k}\cong \Omega_{X/k}$ because they both satisfy the same univeral property. I think if $mK_X$ is very ample, then $X\cong \operatorname{Proj} \bigoplus_{n\geq 0} H^0(X,\omega_{X/k}^{\otimes mn})$, which is Proj of a graded ring generated in degree $1$, so automorphisms of this ring are determined by what happens on the degree-1 piece, or $H^0(X,\omega_{X/k}^{\otimes m})$. Therefore if a map is the identity on $H^0(X,\omega_{X/k}^{\otimes m})$ it must be the identity on $X$.
My issues. I'm pretty unsure about whether my understanding is correct here and I'd like to check my understanding or see a correct proof if mine is flawed. I also have a concern that fact that for a line bundle $\mathcal{L}$, we have $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{L},\mathcal{L})\cong k$. Shouldn't this be saying that there are no interesting maps $f^*\omega_{X/k}\to \omega_{X/k}$, since $f^*\omega_{X/k}\cong \omega_{X/k}$? This seems clearly wrong, but I don't fully understand why (is it possible that maybe $f^*\omega_{X/k}\cong \omega_{X/k}$ is not true as $\mathcal{O}_X$ modules? this is my only guess).
The $ m $ is a red herring anyway: replacing $ mK_X $ by $ K_X $ does not change anything in the following, so we'll assume $ K_X $ itself is very ample.
Anyway, any automorphism $ f $ of $ X $ preserves the canonical bundle. Hence it induces an automorphism of the vector space $ H^0(X, K_X) $ and this automorphism in turn induces an automorphism $ \tilde{f} $ of the projective space $ \mathbb{P}(H^0(X, K_X)^{\vee}) $. Let $ j : X \rightarrow \mathbb{P}(H^0(X, K_X)^{\vee}) $ be the embedding given by the very ample canonical class. There is a commutative diagram $\require{AMScd}$ \begin{CD} \mathbb{P}(H^0(X, K_X)^{\vee}) @>{\tilde{f}}>>\mathbb{P}(H^0(X, K_X)^{\vee}) \\ @AAA @AAA\\ X @>{f}>> X \end{CD}
That is to say the automorphism $ f $ lifts to an automorphism of the ambient projective space given by the canonical embedding. Now it is obvious that $ Aut(X) $ injects into $ Aut(H^0(X,K_X)) $. This argument will also show that $ Aut(X) $ is infact finite if you admit that the field is complex numbers.
Your mistake: it is not true in general that any automorphism of $ X = \operatorname{Proj}(R) $ comes from an automorphism of the degree one piece of $ R $.
Edit, to answer some questions:
(1) For any smooth varieties $ X $ and $ Y $ of equal dimension $ n $ and any morphism $ g : X \rightarrow Y $, the derivative $ dg $ induces a map of tangent bundles $ TX \rightarrow g^* TY $ and hence taking top exterior powers gives a map of line bundles $ \wedge^n dg \in \operatorname{Hom}(\omega_X^{-1}, g^* \omega_Y^{-1}) $. Seeing this as a global section of the line bundle $ \omega_X \otimes g^* \omega_Y^{-1} $, the zero locus of that global section is precisely the points $ x \in X $ where $ dg_x $ fails to give an isomorphism of tangent spaces $ TX_x $ and $ TY_{g(x)} $.
Now specialize to the case when $ Y = X $ and $ g $ is an automorphism in which case $ dg $ is obviously an isomorphism everywhere and hence $ \wedge^n dg $ is nowhere zero. This shows $ \omega_X \otimes g^* \omega_X^{-1} $ is trivial. Therefore any automorphism preserves the canonical bundle.
I didn't really check your way with the universal property of the cotangent bundle but this is my preferred way to see it.
(2) For any line bundle $ L $ on any scheme $ X $, $ \mathcal{H}om(L,L) = \mathcal{O}_X $ so global sections of the structure sheaf give you endomorphisms of $ L $. If $ X $ is projective and $ k $ is algebraically closed, then indeed the global sections are just elements of $ k $. I don't see any contradiction here and I don't really understand what your question is. Obviously all morphisms are morphisms of $ \mathcal{O}_X $-modules, how could they possibly not be here?
(3) For the 'stupidest' example of my last line - which is not mine and I found it in a comment of Jack Huizenga, consider $ X = \mathbb{P}^1 \times \mathbb{P}^1 $ and the embedding given by the very ample bundle $ \mathcal{O}(2,1) $ into $ \mathbb{P}^5 $. Then the obvious involution on $ X $ swaps $ \mathcal{O}(2,1) $ to $ \mathcal{O}(1,2) $ so cannot come from an automorphism of the ambient $ \mathbb{P}^5 $.
(4) For ANY bundle, there is a pullback map of sections. Concretely in this case, there is a pullback of sections $ f^* : H^0(X, \omega_X) \rightarrow H^0(X, f^* \omega_X) $ given by $ s $ mapping to $ s \circ f $. Additionally here we know that $ f $ is an automorphism so the pullback map is a linear automorphism of VECTOR SPACES. This has nothing to do with canonical bundle being preserved, until now because apriori I have never said that $ f^* \omega_X $ is the same as $ \omega_X $.. but I will say it in the next paragraph.
Now aditionally (electric boogaloo) we know there is an isomorphism $ f^* \omega_X \rightarrow \omega_X $ given by $ df $ (which of course must be multiplication by a scalar by point (2)). The induced map on global sections will be called $ \Gamma(df) : H^0(X, f^* \omega_X) \rightarrow H^0(X, \omega_X) $ and this is of course, multiplication by the same scalar df.
The composition $ \Gamma(df) \circ f^* : H^0(X, \omega_X) \rightarrow H^0(X, \omega_X) $ is the map inducing the top arrow.