Can I shorten the resonant length of a tight wire by moving its endpoint in phase?

If I understand you correctly, you want to know what happens when you move one end of a string. Actually this is usually the way you get your standing wave in the first place. Mathematically, you can get the behavior by assuming a time dependent boudary condition at one end, e.g. a $vec A_0sin{omega t}$ time dependence of the amplitude in a certain direction. Then, for transverse $vec A_0$, you will get resonances (standing waves) for angular frequencies corresponding to integer multiples of a half wavelength + a quarter wavelength fitting into the length of the string. For longitudinal $vec A_0$, you will get resonances for angular frequencies corresponding to integer multiples of a half wavelength.

If you have a string oscillating in a standing wave you can quench the oscillation by starting a (longitudinal) sinusoid oscillation with the standing wave frequency at an end point in anti-phase.

Revised Answer (for earlier version click on "edited ... ago")

Reading your question as well as your comments it is still not quite clear what you mean.

I presume that the string of length $L$ with fixed ends A and B is set vibrating, as in the 1st figure below. The tension and mass per unit length determine the speed $v$ of travelling waves. The fundamental frequency is $f=frac{2v}{L}$ (which comes from speed = wavelength x frequency).

P is a point on the string close to end B. It vibrates with frequency $f$. Note that if we force P to vibrate with frequency $f$ then it does not matter whether or not the section PB of the string exists : we could remove it without destroying the standing wave.

With the string at rest end B is then unfixed and forced to vibrate with the same motion that point P had in the 1st diagram. We now have the situation in the 2nd diagram.

If the moving end B has the same frequency $f$ as P, then a standing wave will not be formed on the string. The forcing frequency corresponds to a string of shorter length $L$, but the effective length $L'$ is now somewhat longer than $L$. (How much longer is difficult to say; see the discussion below.) The forcing frequency $f'$ should be correspondingly lower. The tension in the string and its mass per unit length are the same; the speed $v$ of waves on the string is the same. So :
$v = fL/2 = f'L'/2$.

The amplitude of B is not important. Any amplitude will work, but the frequency is crucial.

How much lower than $f$ does $f'$ need to be? I think this cannot be answered exactly. If $L'$ is only slightly greater than $L$, the section BB' will be very steep. This will require the midpoint of the string to have a very large amplitude. A real string may not have enough elasticity to extend so much. To maintain the same amplitude the increase should be approximately proportional :
$f'/f=L/L'=AP/L$.

If P were the midpoint of AB, then $L'=2L$ and $f'=frac12 f$.

This additional answer assumes that the question is about shortening the length of a vibrating string during vibration.

If you are shortening the length of a string that is already vibrating in its fundamental mode, you are moving the resonant frequency to higher values corresponding to the changed length of the string. This is a well-known phenomenon in violins, where during shortening/lengthening the vibrating string length by continuously moving the pressure point on the finger board, a tone with increasing/decreasing pitch is produced.