Because the frequency of a sound wave is defined as "the number of waves per second."
If you had a sound source emitting, say, 200 waves per second, and your ear (inside a different medium) received only 150 waves per second, the remaining waves 50 waves per second would have to pile up somewhere — presumably, at the interface between the two media.
After, say, a minute of playing the sound, there would already be 60 × 50 = 3,000 delayed waves piled up at the interface, waiting for their turn to enter the new medium. If you stopped the sound at that point, it would still take 20 more seconds for all those piled-up waves to get into the new medium, at 150 waves per second. Thus, your ear, inside the different medium, would continue to hear the sound for 20 more seconds after it had already stopped.
We don't observe sound piling up at the boundaries of different media like that. (It would be kind of convenient if it did, since we could use such an effect for easy sound recording, without having to bother with microphones and record discs / digital storage. But alas, it just doesn't happen.) Thus, it appears that, in the real world, the frequency of sound doesn't change between media.
Besides, imagine that you switched the media around: now the sound source would be emitting 150 waves per second, inside the "low-frequency" medium, and your ear would receive 200 waves per second inside the "high-frequency" medium. Where would the extra 50 waves per second come from? The future? Or would they just magically appear from nowhere?
All that said, there are physical processes that can change the frequency of sound, or at least introduce some new frequencies. For example, there are materials that can interact with a sound wave and change its shape, distorting it so that an originally pure single-frequency sound wave acquires overtones at higher frequencies.
These are not, however, the same kinds of continuous shifts as you'd observe with wavelength, when moving from one medium to another with a different speed of sound. Rather, the overtones introduced this way are generally multiples (or simple fractions) of the original frequency: you can easily obtain overtones at two or three or four times the original frequency, but not at, say, 1.018 times the original frequency. This is because they're not really changing the rate at which the waves cycle, but rather the shape of each individual wave (which can be viewed as converting some of each original wave into new waves with two/three/etc. times the original frequency).
I will delay the use of water waves, rather use waves on a string, as in case of water waves, nonlinear effects can be there which can have different dispersion relations like in the case of shallow water/ deep water waves and thereby a different relation for group velocity than you mentioned. So I will derive the case for waves on a string and then sketch the analogous case for water waves, taking into account appropriate approximations. So for waves on a string we have
$$c = \nu \lambda$$
Now we know that the speed of the waves is dependent on the tension and the linear mass density.
$$c = \sqrt{T/\mu}$$ Since neither of them are changing in your example, we know that the speed of the waves is constant. Therefore from the first relation the only thing that changes with change with frequency is the wavelength and not speed.
For water waves, similarly, using appropriate limits if the dispersion relation yields a similar group velocity relation with frequency and wavelength, we can say that since the pressure and the density are not changed, and as a result, the only thing that changes with change with frequency is the wavelength and not speed.
Best Answer
The wavelength of standing waves in a pipe would have to depend on the length of the pipe, since we are dividing the length of the pipe between the standing waves. For example, in the third harmonic, we are dividing the length of the pipe, let's say $L$, into three parts for each 'node-node' pair in the wave, thus giving us $\frac{2L}{3}$ as the wavelength.
This is waves on a string, but the concept is the same.
So, when the density of the particles in the pipe has been changed to change the speed of sound as you said, you may observe that we're still talking about the first harmonic here, which means exactly two antinodes are present on either end of the pipe, and thus the wavelength remains $2L$ as the length of the pipe has not changed. Finally, the frequency will increase by the virtue of wavelength remaining constant while the velocity of the medium has increased.
Your answer key is therefore correct.