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You will notice that the formula is the same as before, except $x$ has been replaced by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.