@Willk Further to my previous answer, and to your comments, let's elaborate with some examples:
Imagine two astronauts floating towards each other in space, one with velocity $v$, the other with $-v$ and both with the same mass $m$. In this reference frame, the total momentum is $mv + m(-v) = 0$ and the total kinetic energy is $mv^2/2 + m(-v)^2/2 = mv^2$ (in classical mechanics, ignoring special relativity for now). If the astronauts now collide (thud!) and grab each other, they will be at rest afterwards. This is because of conservation of momentum: If $u$ is the velocity of both astronauts moving together after the collision, we get $mv + m(-v) = 0 = (m + m)u \implies u = 0$. But now the kinetic energy is $(m + m)u^2/2 = 0$. Because of conservation of energy, the original kinetic energy $mv^2$ must have gone somewhere; it must have been transformed into internal energy in the two astronauts. In this case, most likely into heat in the astronauts' bodies and space suits, but in the case of subatomic particles, it would have been mass (actually, the difference is not clear-cut; as far as relativity is concerned, any internal energy is mass, whether it be heat, internal motion, molecular or atomic binding energy, intrinsic mass coming from the Higgs field...).
So there you are: The kinetic energy of the first astronaut can be converted into internal energy (mass), but only by colliding with something of equal and opposite momentum.
If you're looking for sci-fi writing prompts, the first astronaut could also slow down by ejecting something to carry away all of their momentum, e.g. rocket fuel or a huge blast of radiation. However, this can never result in the astronaut gaining mass, as I'll show below (this time with special relativity, so feel free to skip):
Imagine the astronaut traveling at speed $v$, thus having momentum $p = \gamma mv$ and total (mass + kinetic) energy $E = \gamma mc^2$, where $\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}$. It now ejects something, let us call it radiation, with mass $M$, carrying away all its momentum $p$, so that it comes to a halt (by conservation of momentum). The energy of the radiation is then
$$ \Delta E = \sqrt{(Mc^2)^2 + (pc)^2} \geq pc, $$
and by conservation of energy, the astronaut's total energy afterwards must be $E - \Delta E$. The mass afterwards is just this divided by $c^2$, since the astronaut's momentum is zero:
$$ m' = \gamma m - \frac{\Delta E}{c^2} \leq \gamma m - \frac{p}{c} = \gamma m \left(1 - \frac{v}{c}\right). $$
Abbreviating $\beta = \frac{v}{c}$ and noting that $\beta < 1$ (the astronaut is traveling slower than light), we get
$$ \frac{m'}{m} \leq \gamma(1 - \beta) = \frac{1 - \beta}{\sqrt{1 - \beta^2}} = \sqrt{\frac{1 - \beta}{1 + \beta}} < 1, $$
showing that the astronaut always loses mass.
Since the astronaut (or maybe we should call it a spaceship now) has to collide with something in order to convert its kinetic energy into mass, we might speculate on what... The cosmic microwave background consists of photons coming from all directions in space. So technically, we are all constantly colliding with CMB photons and slowing down relative to the CMB rest frame, but the effect is minuscule. But what if your spaceship had some highly advanced equipment to be able to interact with other things in the environment? Things like dark matter (effect would probably also be minuscule), particles from parallel universes, whatever makes up the vacuum energy of empty space... This is where established physics ends and your creativity starts!