Is there a simple characterization of field extensions $F\subset K$ for which there exists a proper endomorphism (i.e. not surjective), $K\rightarrow K$, fixing $F$ pointwise?
If $K/F$ is algebraic, then no such exists (right?). So I expect the condition to be something about transcendence.
Sure, here are the most basic examples: let $F$ be any field, and let $K$ be a simple transcendental extension, $K=F(t)$. Now, let $g=g(t)$ be any nonconstant element of $K$, thus a rational function with coefficients from $F$. Certainly $g$ is also transcendental over $F$, and we may define $\psi:K\to F(g)\subset K$ by sending $t$ to $g$.
More explicitly, if $f\in F(t)$, let $\psi(f)=f\circ g$. Even more explicitly yet, if $f$ is a polynomial, say $f=\sum_ia_it^i$, then $\psi(f)=f(g(t))=\sum_ia_i(g(t))^i$. And extend $\psi:F[t]\to K$ to the fraction-field in the obvious way. Notice that constants are left fixed by $\psi$.
Simplest example, if $g(t)=t^2$, then $f(t)\mapsto f(t^2)$.
In every case, $[K:\psi(K)]$ is equal to the degree of $g$, where this is just the ordinary degree of $g$ if $g$ is a polynomial, and the max of the degree of the denominator and the degree of the numerator if $g$ is a rational function that’s not a polynomial. ( Of course $g$ must be reduced to lowest terms! )