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The answer is no. By a theorem of Burgoyne, Griess, and me (Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0:=L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $Inn(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all finite overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

The answer is no. By a theorem of Burgoyne, Griess, and me (Maximal subgroups and automorphisms of Chevalley groups, Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0\mathrel{:=}L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $\operatorname{Inn}(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all finite overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

The answer is no. By a theorem of Burgoyne, Griess, and me (Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0:=L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $Inn(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

The answer is no. By a theorem of Burgoyne, Griess, and me (Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0:=L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $Inn(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all finite overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

The answer is no. By a theorem of Burgoyne, Griess, and me (Pacific J. Math. 71 (1977), 365-403), there would exist an integer $f_0$ such that $G_0:=L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $Inn(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

The answer is no. By a theorem of Burgoyne, Griess, and me (Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0:=L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $Inn(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.