A totally-ordered set of functions
This question is very natural since most if not all common functions that appear in the runtime-analysis of algorithms form a totally ordered set in terms of little $o$-notation or big $\Omega$-notation such as shown in this answer by Robert S. Barnes or this answer by Kelalaka.



However, there is no such totally-ordered set of representative functions for all positive increasing functions.

The basic reason is the first property, when applied to $F$, does not define a total-order. 

Define two strictly increasing $f_1(n)$ and $f_2(n)$ from $\Bbb N$ to $\Bbb N$.

$$f_1(n)=2^{(2\lceil\frac {n+1}2\rceil)^2+n}$$
$$f_2(n)=2^{(2\lfloor\frac {n}2\rfloor+1)^2+n}$$

Then 
$$\sup\lim _{n\to\infty}\frac {f_1(n)}{f_2(n)}
\ge \lim_{n=2i+1\to\infty}\frac {f_1(n)}{f_2(n)} 
=\lim_{i\to\infty}\frac {2^{(2(i+1))^2+2i+1}}{2^{(2i+1)^2+2i+1}}
=\lim_{i\to\infty}2^{4i+3}=\infty$$

$$\sup\lim _{n\to\infty}\frac {f_2(n)}{f_1(n)}
\ge \lim_{n=2i\to\infty}\frac {f_2(n)}{f_1(n)} 
=\lim_{i\to\infty}\frac {2^{(2i+1)^2+2i}}{2^{(2i)^2+2i}}
=\lim_{i\to\infty}2^{4i+1}
=\infty$$

In plain words, neither of $f_1(n)$ and $f_2(n)$ grows asymptotically the same or faster than the other one ignoring a constant factor.

Let $F$ and $G$ be defined as in the question. Suppose $G$ exists. Then there is $g_1, g_2\in G$ such that $g_1(n)=\Theta(f_1(n))$ and $g_2(n)=\Theta(f_2(n))$. Then $g_1(n)&gt;c_1(f_1(n))$ for some constant $c_1&gt;0$ and $g_2(n)&lt;d_2(f_2(n))$ for some constant $d_2$. (In usual cases, we have to say "for $n$ large enough". However, that clause can be omitted since all values here are positive. Anyway, we can add that clause as well.)

$$\sup\lim _{n\to\infty}\frac {g_1(n)}{g_2(n)}
\ge \sup\lim_{n\to\infty}\frac {c_1f_1(n)}{d_2f_2(n)} 
\ge \frac{c_1}{d_2}\sup\lim_{n\to\infty}\frac {f_1(n)}{f_2(n)} 
=\infty
$$

Symmetrically, $\sup\lim_{n\to\infty}\dfrac {g_2(n)}{g_1(n)}=\infty$. We see that $g_1\not=g_2$ but neither $g_1(n) = o(g_2(n))$ nor $g_2(n) = o(g_1(n))$. This contradition shows $G$ does not exist.



How about if we drop the first requirement so that the functions in $G$ are not required to dominate each other? 

Then this question becomes closely related to the question Big O notation and the maximal set of comparable functions as found by OP, where a couple of enlightening answers explain the situation pretty well, indicating there is hardly any hope for a nice positive answer. Even countably-infinitely real parameters cannot parametrize continuously a set of representative functions in the equivalence classes of increasing functions where two functions are in the same equivalence class if they are related by $\Theta$ and each function only uses finitely-many parameters. 



Explicit rigorous answers to the following exercises might be long. However, the ideas should be simple for experienced users. 

(Exercise 1.) For any given positive integer $n$, construct $n$ increasing functions from $\Bbb N$ to $\Bbb N$ such that for any two of them, $f_1$ and $f_2$, neither $f_1=O(f_2)$ nor $f_2=O(f_1)$.

(Exercise 2.) Construct a set of infinitely many increasing functions from $\Bbb N$ to $\Bbb N$ such that for any two of them, $f_1$ and $f_2$, neither $f_1=O(f_2)$ nor $f_2=O(f_1)$.

(Exercise 3.) Let $h(n)$ be a given function from $\Bbb N$ to $\Bbb N$. Construct a set of infinitely many increasing functions from $\Bbb N$ to $\Bbb N$ such that for any two of them, $f_1$ and $f_2$, neither $f_1=O(h\ f_2)$ nor $f_2=O(h\ f_1)$.