if $P\_{1}$ and $P\_{2}$ are distinct places of equal degree of the function field F/K, and $\sigma$ is a K-field automorphism, such that $\sigma(P\_{1})=P\_{2}$. then, does $\deg (P\_{1}\cap K(x))=\deg (P\_{2}\cap K(x))$, where K(x) is the rational function field?
in particular, is this true over the hermitian function field? 



behavior of places of a function field under automorphism
No, not in general, that is not without particular requirements for $x$:


take $F=\mathbb{R}(y)$, the rational function field in one variable over the reals.
Then the equation $\sigma (y)=y+1$ determines an automorphism of $F/\mathbb{R}$.


Let $P\_1$ be the place associated to the polynomial $y^2+1$; then $\deg (P\_1)=2$.


Let $P\_2 := \sigma (P\_1)$; then $P\_2$ is associated to the polynomial $y^2+2y+2$ and (automatically) $\deg (P\_2)=2$.


Let $x := y^2+1$; then $[F:\mathbb{R}(x)]=2$ and $P\_1|\_{\mathbb{R}(x)}$ has degree $1$.


On the other hand $yP\_2 $ either equals $i-1$ or $-i-1$. In both cases $xP\_2$ is non-real and thus $\deg (P\_2)=2$.


H


