Here is a diagram with the quantities we need to solve this:
From Kepler's 3rd Law, the "kronostationary" satellite---marked by a green dot---is located at a = 1.86 Req, where Req is the equatorial radius and Rpol is the polar radius (Python calculations dumped at the bottom of the post). The NASA PDS Atmospheres Encyclopedia gives values for the radius of Saturn and useful formulas on its latitude page, and a value for the rotation rate on the longitude page.
At the planetographic latitude ($\phi$g) of the location we want, the line of sight (green) forms a right angle with the local vertical (blue). This sets up a pair of similar triangles as shown. From the big one, we have:
$$ \tan{\phi_{\textrm{g}}} = \frac{a - R_{\textrm{cyl}}}{z} $$
We can use the equation for an ellipse to express z in terms of Rcyl:
$$ \frac{R_{\textrm{cyl}}^2}{R_{\textrm{eq}}^2} + \frac{z^2}{R_{\textrm{pol}}^2} = 1 $$ $$ z = \frac{R_{\textrm{pol}}}{R_{\textrm{eq}}} \sqrt{R_{\textrm{eq}}^2 - R_{\textrm{cyl}}^2} $$
We can then use the pink-red-red triangle to relate $\phi$c and Rcyl, eliminating z using the equation above:
$$ \tan{\phi_{\textrm{c}}} = \frac{z}{R_{\textrm{cyl}}} = \frac{R_{\textrm{pol}}}{R_{\textrm{eq}} R_{\textrm{cyl}}} \sqrt{R_{\textrm{eq}}^2 - R_{\textrm{cyl}}^2} $$
The PDS Atmospheres formulas relate $\phi$g and $\phi$c:
$$ \tan{\phi_{\textrm{c}}} = \left(\frac{R_{\textrm{pol}}}{R_{\textrm{eq}}}\right)^2 \tan{\phi_{\textrm{g}}} $$
Combining the first and last equations, and eliminating z, gives a surprisingly simple equation:
$$ R_{\textrm{cyl}} = \frac{R_{\textrm{eq}}^2}{a} = 32362\ \textrm{km} $$
Plugging this value of Rcyl into the equations above gives: - Planetocentric latitude 54.8° - Planetographic latitude 60.1°
Actual visibility and survivability of this satellite are not great. The kronostationary radius would put this satellite within the B ring, which is "the largest, brightest, and most massive of the rings:"
Calculations:
import numpy as np
R_eq = 60268. # km
R_pol = 54364. 371
# km
GMsat = 37931206.159 # km^3/s^2
P_sat = 10.656 * 60.**2. # s
# <PERSON><PHONE_NUMBER> # km^3/s^2
P_sat = 10.656 * 60.**2. # s
# Kepler's 3rd law: a is the kronocentric distance (circular orbit)
a = (P_sat**2 * GMsat / 4 / (np.pi)**2)**(1/3.)
print("Kronocentric distance,")
print(" a (R_sat): ", a / R_eq)
print(" a (km): ", a)
print(" altitude (km): ", a - R_eq)
print(" ")
# solve for R_cyl
R_cyl = R_eq**2. / a
print("R_cyl: ", R_cyl)
print(" ")
# get latitude
lat_centric = np.arctan(R_pol / R_eq / R_cyl * (R_eq**2. - R_cyl**2.)**0.5)
lat_graphic = np.arctan(R_eq**2. / R_pol**2. * np.tan(lat_centric))
print("lat_centric: ", lat_centric * 180./np.pi)
print("lat_graphic: ", lat_graphic * 180./np.pi)
print(" ")
Kronocentric distance,
a (R_sat): 1. 371