This is not a complete answer: it is instead an attempt to provide the information needed for one. In particular I will answer the question 'how fast did the exposure need to be to take that picture' and show that the answer is 'not stupidly fast', and argue that you don't need to do anything heroic to record such an image with a commercial sensor. I think this is all that can really be said without a detailed spec sheet for the sensor, which I could not find.
So first of all we know the blade's rotational velocity: $2600\,\mathrm{rpm}$. Turning this into radians per second gives
$$
\begin{aligned}
\omega &= \frac{2600 \times 2\pi}{60}\\
&\approx 272\,\mathrm{s^{-1}}
\end{aligned}$$
The total length of the blades is $1.2\,\mathrm{m}$ so each blade is half that, which gives a tip velocity of
$$v_\mathrm{tip} \approx 163\,\mathrm{ms^{-1}}$$
This smells good as it's subsonic but not vastly so (speed of sound on Mars is $\sim 240\,\mathrm{ms^{-1}}$), and we know keeping the tips comfortably subsonic was a consideration.
But what we actually want is the number of pixels per second. Well, we know the image is 640 by 480 pixels, and if you measure the vertical pair of blades they are pretty accurately half the height (480 pixels) of the image. So that gives a single blade length $l_\mathrm{px} \approx 120\,\mathrm{px}$. So now
$$
\begin{aligned}
v_\mathrm{tip,px} &= \omega l_\mathrm{px}\\
&\approx 32600\,\mathrm{px/s}
\end{aligned}
$$
So, now, that means to get an image where the tip has moved less than one pixel, you need an exposure of about $1/32600\,\mathrm{s}$.
So then one question is: is the image anything like that sharp? Well, I fetched the best version I could, from here (and I compared it with several other version such as the Wikipedia one: they all look the same). And the answer is 'no': the whole image is no sharper than 2-3 pixels I should think, although it is possible that this is a processed image, in which case the original would be more interesting (if it ever left Ingenuity, which it may not have!). So I think, conservatively, we could divide by two here: an exposure of $1/16300\,\mathrm{s}$ would easily give the image we see, and quite possibly one of $1/8000\,\mathrm{s}$ would be fine.
So I will assume, finally, $1/16000\,\mathrm{s}$ as my best guess as a safe upper bound for the length of the exposure (note I have rounded it again).
So there are now two questions to which I don't have an answer, but I can make what I think are good guesses. Both of these are assuming that the camera sensor is an OmniVision OV7251. Unfortunately the spec sheets I could find for this device were not helpful enough for me to provide definitive answers to the below for this specific sensor.
Is such a fast shutter speed possible? Yes, easily. This is an electronic shutter, it has no moving parts, it's almost certainly limited only by getting enough photons into the sensor while it's open. Note that the shutter speed is not the same thing the inverse of the frames-per-second rate of the device: that rate is how fast it can capture an image, read it from the sensor, do whatever initial processing is needed, and dump it down the interconnect. The frames-per-second rate gives an upper bound on the length the shutter can be open but no lower bound. As an aside: I have a film camera with an electronically-controlled mechanical shutter which can do $1/4000\,\mathrm{s}$ (but this would be a moving slit, which the OV7251 explicitly is not).
Which brings us to the second, more interesting question.
Is the sensor sensitive enough to capture images at this shutter speed? Well, evidently the answer to this is 'yes', but I could not find any information on the sensitivity of the OV7251 except a rather vague statement about 'excellent low-light sensitivity'.
However I think it is very plausible that this is easily possible. Unfortunately I don't know what the right measure for sensor sensitivity is outside photographic use. But modern digital sensors can be absurdly sensitive: some modern digital cameras are perfectly usable at ISOs of over 200,000. The pixels on this sensor ($3\,\mathrm{\mu m}$, it says) are fairly small (bigger pixels are better for high-sensitivity if you want images with low noise), but on the other hand the image quality from it is not great, quite apart from resolution. Of course it doesn't have to be great: it's a navigation camera, not a science camera.
And I think it's also plausible that, given the likely target market for this sensor, it's going to be pretty sensitive: if it's being used for things like motion detection or object detection about the last thing you want from it is an artistically motion-blurred image of the thing you're about to crash into: you want the image as sharp as it can be. And you want to be able to do this at night.
Guessing the ISO with sunny 16
I realised overnight it's possible to make a guess at the effective ISO of the sensor, using the famous 'sunny 16' rule known to anyone who has used a camera without a lightmeter.
The sunny 16 rule. In bright sunlight, on Earth, a reasonable exposure for a film with ISO sensitivity $S$ is $1/S$ at $f/16$.
Well, we can generalise this. If the aperture of the lens is $A$ (so it's an $f/A$ lens) then, on Earth the exposure you need in bright sunlight is
$$t(A) = \left(\frac{A}{16}\right)^2 \frac{1}{S}$$
But we're no on Earth, we're on Mars, and the amount of light that arrives goes as the inverse square of the orbital radius, so:
$$t(A,R) = \left(\frac{A R}{16 R_E}\right)^2 \frac{1}{S}$$
Where $R$ is the orbital radius of whichever planet we are on, and $R_E$ is Earth's orbital radius.
Or in other words
$$S = \left(\frac{A R}{16 R_E}\right)^2 \frac{1}{t}$$
so if we know $t$ we can guess $S$, if we know $A$.
So let's try for three possible apertures
- $f/2$ – fast, but not absurd, probably would make focusing interesting when Ingenuity was on the ground even though the sensor is small though;
- $f/4$ – perhaps more plausible;
- $f/16$ – slow, small, cheap.
And we know that $R_M/R_E \approx 1.5$, and based on the image the sin is difectly overhead (and I am ignoring atmospheric conditions like inconvenient clouds, because probably they would not fly anyway if there was not enough sunlight to recharge with). And I'll use my safe guess of $t = 1/16000\,\mathrm{s}$.
- with an $f/2$ lens we get $S = (2\times 1.5/16)^2\times 16000 \approx 560$;
- with an $f/4$ lens we get $S = (4\times 1.5/16)^2\times 16000 \approx 2250$;
- with an $f/16$ lens we get $S \approx 36000$.
The first two sensitivities would be achievable with film, the last is very easily within the reach of a commercial digital sensor.
In summary:
- capturing this image probably required an exposure of no longer than $1/16000\,\mathrm{s}$, although this is a rather conservative estimate;
- an exposure this short should be easily achievable with an electronic shutter, but without detailed specifications of the shutter on the OmniVision OV7251 sensor I don't know for sure;
- using 'sunny 16' I can guess the required sensitivity of the sensor in ISO, and the values are not extreme, and, if the lens is fast, rather low.
