Just follow your nose. Functoriality of cohomology gives a map $\operatorname{Aut}(X_\bar{\mathbb{Q}})\to \operatorname{Aut}\big(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)\big)$ (actually, the functoriality is contravariant, so gives a map to the opposite group -- but for groups, $G \cong G^{op}$ via $g\mapsto g^{-1}$). This map is compatible with Galois action, hence you get (again by functoriality) a map $$H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(X_{\bar{\mathbb{Q}}})\big)\to H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}))\,\big).$$ The RHS once again classifies Galois twists (now of a vector space).
There is a topological way of seeing why this functoriality should hold. Very informally, you should think of an equivariant object (e.g. $X_{\bar{\mathbb{Q}}}$) of any category with $\Gamma$ action (for $\Gamma$ the Galois group) as a local system of objects in a local system of categories over a topological space $S$ with $\pi_1(S) = \Gamma.$ The Galois action $\Gamma\to \operatorname{Aut}(X)$ is encoded by the monodromy; a twist in $H^1(S, \operatorname{Aut})$ is some class over $X$ which can twist the monodromy. Now any functor of $\Gamma$-equivariant categories gives maps of all the relevant data over $X$, and in particular takes a twist to a twist.