They are not. $H^0(\Gamma, \text{Aut}(X_{\bar{k}}))$ computes the subgroup of automorphisms which are Galois-invariant, which is equivalently the automorphism group of $X$, and similarly for $Y$, so to find a counterexample it suffices to find varieties $X, Y$ which are isomorphic over $\bar{k}$ but which have non-isomorphic automorphism groups over $k$.

Let $k = \mathbb{R}$, let $X = \mathbb{P}^1$, and let $Y$ be the conic $\{ X^2 + Y^2 + Z^2 = 0 \}$ in $\mathbb{P}^2$. Then $\text{Aut}(X) \cong PGL_2(\mathbb{R})$ but $\text{Aut}(Y) \cong PO(3) \cong SO(3)$, and these groups can be distinguished abstractly by their abelianizations: the abelianization of $PGL_2(\mathbb{R})$ is $\mathbb{R}^{\times}$ (given by the determinant), while the abelianization of $SO(3)$ is trivial.

They are not. $H^0(\Gamma, \text{Aut}(X_{\bar{k}}))$ computes the subgroup of automorphisms which are Galois-invariant, which is equivalently the automorphism group of $X$, and similarly for $Y$, so to find a counterexample it suffices to find varieties $X, Y$ which are isomorphic over $\bar{k}$ but which have non-isomorphic automorphism groups over $k$.

Let $k = \mathbb{R}$, let $X = \mathbb{P}^1$, and let $Y$ be the conic $\{ X^2 + Y^2 + Z^2 = 0 \}$ in $\mathbb{P}^2$. Then $\text{Aut}(X) \cong PGL_2(\mathbb{R})$ but $\text{Aut}(Y) \cong PO(3) \cong SO(3)$ (actually I'm not entirely confident I know how to prove this, at least not without passing to the complexification). These two groups can be distinguished abstractly by their abelianizations: I believe (but haven't checked carefully) that the abelianization of $PGL_2(\mathbb{R})$ is $\{ \pm 1 \}$ (given by the sign of the determinant) but the abelianization of $SO(3)$ is trivial.