Let $X$ be a rationally connected, smooth projective variety over $\mathbb{C}$. Then let $Aut^{0}(X)$ be the identity component of the automorphism group.
I would like to find a reference or a proof for the following fact:
Fact: $Aut^{0}(X)$ is a linear algebraic group.
$\def\Aut{\mathrm{Aut}}\def\Pic{\mathrm{Pic}}\def\PGL{\mathrm{PGL}}\def\GL{\mathrm{GL}}\def\SL{\mathrm{SL}}$ When I first read the question, I thought the request was for a reference for the fact that $\Aut^0(X)$ is an algebraic group, which is true without the hypothesis of rational connectedness. The usual source to cite for this is
Matsusaka, T., Polarized varieties fields of moduli and generalized Kummer varieties of polarized abelian varieties, Am. J. Math. 80, 45-82 (1958).
On a second read, I think the OP knows this and is looking for a citation for the fact that, if $X$ is rationally connected, then $\Aut^0(X)$ is a linear algebraic group (and not, say, an elliptic curve). I don't know a reference for this, but I can explain it. The proof falls into two parts: (1) Rationally connected implies $\Pic^0(X)$ is trivial and (2) $\Pic^0(X)$ is trivial implies that $\Aut^0(X)$ embeds in $\PGL$ of a large projective embedding. (Recall that $\PGL_n$ is a linear algebraic group. Proof: For $g \in \GL(V)$, let $g^{\otimes k}$ be the matrix by which $g$ acts on $V^{\otimes k}$. Then $\PGL_n$ embeds in $\GL_{n^n}$ by $g \mapsto g^{\otimes n}/(\det g)$.)
If $X$ is a rationally connected, then $\Pic^0(X)$ is trivial. From the exponential sequence $1 \to \mathbb{Z} \overset{2 \pi i}{\longrightarrow} \mathcal{O} \overset{\exp}{\longrightarrow} \mathcal{O}^{\ast} \to 1$, we identify $\Pic^0(X)$ with the cokernel of $H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O})$, so it is enough to show that $H^1(X, \mathcal{O})=0$. By symmetry of the Hodge diamond, it is equivalent to show that $H^0(X, \Omega^1)=0$.
The condition $H^0(X, \Omega^1)=0$ is a standard property of rationally connected varieties. A reference is Corollary 3.8 in Kollar's book Rationally Connected Varieties (1996).
I tried to write out my own proof of $H^0(X, \Omega^1)=0$ from scratch (in particular, without Kollar's Theorem 3.7) and found it harder than I expected: Here is what I came up with. Suppose that $\omega$ is a nonzero global $1$-form on $X$. By Hodge theory, $d \omega=0$. Thus, $\int \omega$ is a well-defined linear functional on $H_1(X, \mathbb{Z})$, and therefore $\int_{\gamma} \omega$ can only take countably many values as $\gamma$ ranges over $1$-cycles. If $\omega$ is nonzero, then we can find two points $p$ and $q$ and a path $\alpha$ from $p$ to $q$ where $\int_{\alpha} \omega$ is not one of these countably many values. Now, take a rational chain linking $p$ and $q$ and let $\beta$ be a path from $q$ to $p$ within this chain. Since every $1$-form on a rational curve is $0$, we have $\omega|_{\beta} =0$ and $\int_{\beta} \omega =0$. So $\int_{\alpha + \beta} \omega = \int_{\alpha} \omega$. But $\alpha+\beta$ is a closed cycle, contradicting our condition that $\int_{\alpha} \omega$ is not a value of $\int_{\gamma} \omega$ for cycles. $\square$
$\Pic^0(X)$ is trivial implies that $\Aut^0(X)$ embeds in $\PGL$: Let $L$ be a very ample line bundle, so $X$ embeds naturally in $\mathbb{P}(H^0(X, L)^{\vee})$. For any $g \in \Aut(X)$, we have a natural diagram $$\begin{matrix} X & \overset{g}{\longrightarrow} & X \\ \downarrow & & \downarrow \\ \mathbb{P}(H^0(X, g^{\ast} L)^{\vee}) & \overset{(g^{\ast})^{\vee}}{\longrightarrow} & \mathbb{P}(H^0(X, L)^{\vee}) \\ \end{matrix} .$$
Now, since $\Pic^0(X)$ is trivial, $\Pic(X)$ is discrete and $\Aut^0(X)$ acts trivially on $\Pic(X)$. So $L \cong g^{\ast} L$. This isomorphism is not unique but is unique up to multiplication by a scalar. (In general, it would be unique up to an element of $\Aut(L)$, which is to say a nonvanishing global function on $X$, but $X$ is connected and projective, so the only such functions are scalars.) Hence, the isomorphsm $\mathbb{P}(H^0(X, L)^{\vee}) \to \mathbb{P}(H^0(X, g^{\ast} L)^{\vee})$ induced by $g^{\ast} L \to L$ IS unique. Adding this to the digaram above, we get $$\begin{matrix} X &=& X & \overset{g}{\longrightarrow} & X \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{P}(H^0(X, L)^{\vee}) & \cong & \mathbb{P}(H^0(X, g^{\ast} L)^{\vee}) & \overset{(g^{\ast})^{\vee}}{\longrightarrow} & \mathbb{P}(H^0(X, L)^{\vee}) \\ \end{matrix} .$$
The bottom maps give an action of $g$ on $\mathbb{P}(H^0(X, L)^{\vee})$ (I leave it to you to check that it is an action) and hence a map $\Aut^0(X) \to \PGL_N$. We need to check that this map is injective. Suppose that $g$ acts trivially on the bottom row. Since the vertical maps are injections, it must act trivially on the top row. But this is the definition of $g$ being trivial in $\Aut^0(X)$. $\square$