# Does composing the Frobenius with an automorphism give another Frobenius

Let $X_0$ be a variety over $\mathbf{F}_q$. Consider the Frobenius $F_0:X_0\to X_0$. Let $X= X_0\times \bar{\mathbf{F}_q}$ and let $F:X\to X$ be $F_0 \times \textrm{id}$.

Let $f:X\to X$ be an automorphism of finite order. Is $F\circ f$ the Frobenius with respect to some new way of lowering the base field to $\mathbf{F}_{q}$?

• Could you make more precise your question ? – user18119 Feb 22 '12 at 12:29