Temperature depends on the average kinetic energy of the particles of the system. If a system is moving, it possesses kinetic energy, which means the particles also possess that kinetic energy due to the motion of the system; hence, the temperature of the system rises.
Can I please receive confirmation for my reasoning as stated above?
No, the thermal energy of the particles in a rigid body (i.e., the vibrations of atoms about their equilibrium points) is an internal property. This thermal motion is defined in the rest frame of the system, the frame in which the center of mass of the body is at rest.
If the whole body moves with velocity $v$, then all the atoms move together with the same velocity relative to the lab frame. But from the perspective of the body's rest frame, the atoms are still at rest, and no internal motion is triggered. A frame moving at constant velocity $v$ is physically equivalent to one at rest (Galilean invariance in classical mechanics).
So, the kinetic energy associated with the system's bulk motion is not counted as thermal energy, and the temperature remains unchanged. Temperature depends only on the random internal motion of particles, not on the overall motion of the system.
Does a moving body possess a greater temperature than when it is at rest?
No. Short proof below.
Temperature of a particle system according to the kinetic theory of gases is proportional to the average molecule speed squared, i.e. $T \propto {\bar v}^{^2}$. Let's try to calculate average molecule speed wrt to outside observer, which is,
$$\tag 1 \bar v = \frac1N \sum_{i=1}^N (v_i+v_{COM}\cos\theta_i)$$
where $N$ is molecules amount in a gas container, $v_{COM}$ is container speed according to some external observer and $\theta_i$,- angle between i-th molecule own velocity vector about COM (center of mass in gas container) and COM velocity vector itself.
Second term under summation in (1) is gas container COM speed scalar projection into individual molecule speed $v_i$. This is needed to do, because kinetic theory of gases deals only with individual molecule speeds and so we have adjusted molecule individual speed to account for a global system movement.
(1) can be split into two symmetric parts, $$ \bar v = \frac1N \sum_{i=1}^N v_i + v_{COM}\frac1N \sum_{i=1}^N \cos\theta_i \tag 2$$
We notice that term under second summation is just an expected value of $\cos\theta$, over uniform random distribution of $\theta \in [0..2\pi]$ (because molecules moves randomly). So (2) can be stated as,
$$ \tag 3 \bar v = \frac1N \sum_{i=1}^N v_i + v_{COM} \mathbb E[\cos \theta] $$
By the theory of moments in the continuous uniform distribution , $\cos \theta$ mean in the (3) can be re-stated with the help of definite integral, as so :
$$ \tag 4 \bar v = \frac1N \sum_{i=1}^N v_i + v_{COM} \left(\frac {1}{2\pi} \int_0^{2\pi}\cos \theta ~d\theta \right)$$
Integral in the (4) results in zero, and so we arrive at the standard average speed definition,
$$ \tag 5 \bar v = \frac1N \sum_{i=1}^N v_i $$
Which shows that average speed of molecules is invariant (independent) of system COM uniform and linear motion. Hence, whatever external reference frame you will choose,- as long as box is moving at a constant velocity $\vec v_{COM}$,- all observers will detect same gas temperature as in gas own reference frame, i.e. $T'=T.$
