a random variable is understood as a measurable function defined on a probability space whose outcomes are typically real numbers.
every function has its domain.
I assume the domain of standard normal distribution is $(-\infty, \infty)$ (wiki does not claim this, so, more solid References may be needed)
what wiki does claim is
Every normal distribution is a version of the standard normal distribution whose domain has been stretched by a factor ${\displaystyle \sigma }$ (the standard deviation) and then translated by ${\displaystyle \mu }$ (the mean value)
considered the stretch and translation, Do general normal distribution and standard normal distribution have the same domain or not?
Best Answer
As explained in the comments, you are using slightly unusual terminology which is causing some of your confusion.
In fact, I would suggest you to forget about random variables for this question - it is not adding anything and only confusing things. Instead, your question is about the standard normal distribution, which can be rigorously defined as the object which assigns to each interval $(a,b)\subseteq \mathbb R$ the number $$ \frac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2}\ dx. $$ Notice there is no random variable involved in this definition. (The number assigned to the interval $(a,b)$ can be interpreted as the probability that a standard normal random variable lies in $(a,b)$ - but I would recommend thinking of this as a consequence of the definition.)
Saying that the "domain" (or more precisely, "support") of this distribution is $\mathbb R$ simply means that all non-empty intervals $(a,b)$ are assigned a strictly positive number by the above formula. In fact, this is easy to see since you can get a lower bound for the integral by using the minimum of the density (which always occurs at one of the endpoints), and integrating a positive number over an interval of positive size yields another positive number. Common examples of distributions where this is not the case would be the Uniform$[a,b]$ distribution, the Exponential$(\lambda)$ distribution, or any discrete distribution - in each of these cases, you can find an interval that gets assigned $0$ - for instance, $(a-2,a-1)$ in the first case, $(-2,-1)$ in the second case, and any interval that avoids the discrete values in the third case.
Now I will clarify what is meant when it is said that every normal distribution is obtained by translating and stretching a standard normal distribution. The normal distribution with mean $\mu$ and variance $\sigma^2$ can be defined rigorously as the object which assigns to the interval $(a,b)$ the number $$ \frac{1}{\sqrt{2\pi}}\int_a^b e^{-(\frac{x-\mu}{\sigma})^2/2}\ dx. $$ You will notice that the formula is the same as before, except $x$ has been replaced by $$ \frac{x-\mu}{\sigma}. $$ If you think about what the operation $x\mapsto (x-\mu)/\sigma$ does, you will see that it has the effect of shifting the real line $\mathbb R$ over by $\mu$, and then stretching by a factor $\sigma$. However, this does not mean that the domain itself is changing: all that has happened is that the domain has been relabeled, but as a whole it has not changed.