Claudio's answeranswer settles the determination of $\mathrm{Aut}^{\mathrm{o}}(F)$$\DeclareMathOperator\Aut{Aut}\Aut^{\mathrm{o}}(F)$; that of $\mathrm{Aut}(F)$$\Aut(F)$ is more subtle. For Grassmannians this is a classical result of Chow (AnnOn the geometry of algebraic homogeneous spaces, Ann. of Math. (2) 50 (1949), 32-67): $\mathrm{Aut}(\mathbb{G}(p,n))$$\Aut(\mathbb{G}(p,n))$ is connected (hence equal to $\mathrm{GL}(n)$$\operatorname{GL}(n)$) except in the case $n=2p$, where $\pi _0(\mathrm{Aut}(\mathbb{G}(p,2p ))=\mathbb{Z}/2$$\pi _0(\Aut(\mathbb{G}(p,2p ))=\mathbb{Z}/2$. This has been extended to all flag varieties by Tango: On the automorphisms of flag manifolds. Bull. Kyoto Univ. Ed. Ser. B No. 49 (1976), 1-91–9. He proves that the automorphism group of $\mathbb{F}(n;n_1,\ldots ,n_k)$$\mathbb{F}(n;n_1,\dotsc ,n_k)$ is connected except in the case $n_i+n_{k-i}=n$ for all $k$, where it has 2 components.