For reference, here is table 2 from page 4 in reference [$1$]:
- When we reconvert an SI unit back into an SI unit, shouldn't we get back 1 kg?
Yes. The source of confusion here seems to be a misunderstanding of what the first line in table 2 means. It means that in a system of units in which the speed of light ($c$) and Newton's constant ($G$) are both equal to $1$, the SI unit "$1$ meter" can also be expressed as a number of kilograms, specifically $1.3466\times 10^{27}$ kg. The table is not starting with $1$ kg and then ending up with $1.3466\times 10^{27}$ kg. Instead, it is showing how to express "$1$ meter" in kilograms. Here's the explicit calculation, using $c=G=1$, keeping only two significant digits for simplicity:
\begin{align}
1\text{ meter}
&= 1\text{ m }\times\frac{c^2}{G} \\
&\approx 1\text{ m }\times\frac{
(3.0\times 10^8\text{ m/s})^2
}{
6.7\times 10^{-11}\text{ m}^3/(\text{kg}\cdot\text{s}^2)
} \\
&\approx 1\text{ m }\times\frac{1.3\times 10^{27}\text{ kg}}{1\text{ m}} \\
&\approx 1.3\times 10^{27}\text{ kg}.
\end{align}
The first step simply multiplies by $1$, expressed as $c^2/G$. Since we're using units in which $c=G=1$, we could also multiply by $1$ expressed as, say, $c^{42}G^{7/3}$, if we wanted to, because that's also equal to $1$ in these units, and the result would still be legitimate. However, that would give an awkward combination of the SI units "meter" and "kilogram" on the right-hand side. The reason for multiplying by $c^2/G$ is that all of the meters cancel, leaving only kilograms, so we can use this to express a given number of meters as some number of kilograms, or conversely. For example, the mass $M$ of the sun is $2.0\times 10^{30}$ kg, which can be expressed in meters like this:
\begin{align}
M\approx 2.0\times 10^{30}\text{ kg}
&\approx 2.0\times 10^{30}\text{ kg}\times
\frac{1\text{ m }}{1.3\times 10^{27}\text{ kg}} \\
&\approx 1.5\times 10^3\text{ m}.
\end{align}
The Schwarzschild radius of the sun is $R=2GM/c^2$, which can be written simply as $R=2M$ in units where $c=G=1$. Either way, it comes out to be $R\approx 3$ km.
- On pg. 2, they say $E$ can be any unit of energy. How is that? Clearly assuming $E=1$ J and $E=1$ GeV gives us different answers.
Of course, $1$ Joule and $1$ GeV are two entirely different amounts of energy. They are not equivalent. On the contrary, $1$ GeV is equivalent to $\approx 1.6\times 10^{-10}$ Joules, according to page 126 in reference [2] and also acknowledged explicitly on page 2 in reference [$1$]. Page 2 in reference [$1$] is saying that if we use units in which $c$ and Planck's constant $\hbar$ are both equal to $1$ (that is, $c=\hbar=1$), then we can express kilograms, meters, and seconds all in units of energy. Once we have a quantity in units of energy, we can express it either in GeV or in Joules — with different numeric values, of course, because $1$ GeV $\approx 1.6\times 10^{-10}$ Joules. That's all reference [$1$] means by "where $E$ is an arbitrarily chosen energy unit" below equation (2); nothing novel, just the usual freedom to express a given amount of energy using either GeV, or Joules, or ergs ($1$ erg $=10^{-7}$ J), or kilowatt-hours, or whatever happens to be convenient.
References:
[$1$] Myers, "NATURAL SYSTEM OF UNITS IN GENERAL RELATIVITY," https://www.seas.upenn.edu/~amyers/NaturalUnits.pdf
[2] "The International System of Units (SI), 8th edition," International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf
