Was wondering this while running over lunch today. Strange question I know but the mind does wander while running. Made me curious. I’d guess running into a 50 mph wind would be like running in a pool up to your neck.
It depends on the velocity you are running.
To take an example “running” at 0mph in water is equivalent to a 0mph wind, but running at 10mph in water is certainly not the same as running into a 10mph wind.
I did a quick calculation and came up with a difference of 800x (you’d have to go 800x faster to have the same air drag as water drag), but that would have many caveats: you’d need to be completely underwater, both fluids would have to be in the speed range for turbulent flow, and there are probably other limitations. It wouldn’t be simple to get a perfect calculation.
My guess is that 50mph winds would be less strong than running a pool up to your neck. You’d probably decelerate at like 1g on the bike, but if you suddenly found yourself in a pool up to your neck, even doing a couple mph, you’d probably decelerate even faster, hinting that the forces are bigger in the pool.
I think there are too many transitions to make a linear approximation. Running at 2mph in a pool would be very different than being in a 1600mph wind (at that point we are dealing with supersonic dynamics).
When the wind is so strong that you go faster by swimming horizontally instead of running.
It can not be equivalent because of bouyancy and cavitation. Also, your limb speed in water would be the same order of magnitude as your core speed with respect to the fluid, but in high speed air it would be proportionally much smaller, so dynamic similarity fails.
Water tunnels can be used to model certain flows in air as long as key similarity parameters are handled well. Astronauts train for space missions underwater, so maybe it simulates “no air” best 
I think there are too many transitions to make a linear approximation. Running at 2mph in a pool would be very different than being in a 1600mph wind (at that point we are dealing with supersonic dynamics).
Forgetting the rest of the physics the appropriate ratio would be found as follows
D_w = D_a
1/2rho_wV_w^2Cd_wA=1/2rho_aV_a^2Cd_aA
If you assume that the drag coefficient is the same (it likely can’t be)
V_a= sqrt(rho_w/rho_a)*V_w is approximately sqrt(816)*V_w
For 2mph V_w, and some likely incorrect assumptions about drag coefficients, equivalent drag velocity in air V_a = 57 mph.
I think this would be close as per this
I think there are too many transitions to make a linear approximation. Running at 2mph in a pool would be very different than being in a 1600mph wind (at that point we are dealing with supersonic dynamics).
Forgetting the rest of the physics the appropriate ratio would be found as follows
D_w = D_a
1/2rho_wV_w^2Cd_wA=1/2rho_aV_a^2Cd_aA
If you assume that the drag coefficient is the same (it likely can’t be)
V_a= sqrt(rho_w/rho_a)*V_w is approximately sqrt(816)*V_w
For 2mph V_w, and some likely incorrect assumptions about drag coefficients, equivalent drag velocity in air V_a = 57 mph.