The short answer is that it can, if $M = 1 = M^{-1}$. In this way of looking at it, all quantities in Planck units are pure numbers.

The longer answer is that there are two different ways of thinking about natural unit systems.

# Natural unit systems in terms of standard units

One of them, and perhaps the easier one to understand, is that you're still working in a "traditional" unit system in which distinct units for all quantities exist, but the units are chosen such that the numerical values of certain constants are equal to 1. For example, if you want to set $c = 1$, you're not *literally* setting $c = 1$, you're actually setting $c = 1\,\frac{\text{length unit}}{\text{time unit}}$. Length and time don't actually have the same units in this interpretation; they're equivalent up to a multiplication by factors of $c$. In other words, it's understood that to convert from, say, a time unit to a length unit you multiply by $c$, and so that is left implicit.

In order to do this, of course, you have to choose a length unit and time unit which are compatible with this equation. So you couldn't use meters as your length unit and seconds as your time unit, but you could use light-seconds and seconds, respectively.

If you want to set multiple constants to have numerical values of 1, that constrains your possible choices of units even further. For example, suppose you're setting $c$ and $G$ to have numerical values of 1. That means your units have to satisfy both the constraints

$$\begin{align}
c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_G}{t_G} &
G &= \frac{(\text{length unit})^3}{(\text{mass unit})(\text{time unit})^2} = \frac{\ell_G^3}{m_Gt_G^2}
\end{align}$$

where I've introduced $\ell_G$, $t_G$, and $m_G$ to stand for the length, time, and mass units in this system, respectively. You can then invert these equations to solve for $\ell_G$, $t_G$, and $m_G$ in terms of $c$ and $G$ - but as you can probably tell, the system of equations is underdetermined. It still gives you the freedom to choose one unit to be part of your unit system, such as

$$\text{kilogram} = \text{mass unit} = m_G$$

Having made that choice, you can now solve for $m_G$, $\ell_G$, and $t_G$ in terms of $c$, $G$, and $\text{kilogram}$ (or whatever other choice you might have made; each choice gives you a different unit system).

Running through the math for this gets you

$$\begin{align}
m_G &= 1\text{ kg} &
\ell_G &= \frac{G (1\text{ kg})}{c^2} &
t_G &= \frac{\ell_G}{c} = \frac{G (1\text{ kg})}{c^3}
\end{align}$$

Now you can plug in values of $G$ and $c$ in, say, SI units, and get conversions from SI (or whatever) to this unit system. Note that, as I said, length does not *literally* have the same units as time or mass, but you can convert between the length unit, time unit, and mass unit by multiplying by factors of $G$ and $c$, constants which have numerical values of 1. In a sense, you can consider this multiplication by $G^ic^j$ as analogous to a gauge transformation, i.e. a transformation that has no effect on the numerical value of a quantity, and the units of length, time, and mass are mapped on to each other by this transformation just as gauge-equivalent states are mapped on to each other by a gauge transformation in QFT. So it's more proper to say $L \sim T \sim M$; the dimensions are not *equal*, just *equivalent* under some transformation.

If you do the same thing but setting $c = \hbar = 1$ instead, remember what you're really doing is specifying that your units must satisfy the constraints

$$\begin{align}
c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_Q}{t_Q} &
\hbar &= \frac{(\text{length unit})^2(\text{mass unit})}{(\text{time unit})} = \frac{\ell_Q^2m_Q}{t_Q}
\end{align}$$

($Q$ is for "quantum" because these are typical QFT units), and then running through the math, again with $m_Q = 1\text{ kg}$, you get

$$\begin{align}
m_Q &= 1\text{ kg} &
\ell_Q &= \frac{\hbar}{(1\text{ kg})c} &
t_Q &= \frac{\ell_Q}{c} = \frac{\hbar}{(1\text{ kg})c^2}
\end{align}$$

Again, the units are not literally identical, but $\ell_Q \sim t_Q \sim m_Q^{-1}$ under multiplication by factors of $\hbar$ and $c$.

Of course, your third constraint doesn't *have* to be a choice of one of the fundamental units. You can also choose a third physical constant to have a numerical value of 1. To obtain Planck units, for example, you would specify

$$\begin{align}
c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_P}{t_P} \\
\hbar &= \frac{(\text{length unit})^2(\text{mass unit})}{(\text{time unit})} = \frac{\ell_P^2m_P}{t_P} \\
G &= \frac{(\text{length unit})^3}{(\text{mass unit})(\text{time unit})^2} = \frac{\ell_P^3}{m_Pt_P^2}
\end{align}$$

You can tell that this is no longer an underdetermined system of equations. Solving it gives you

$$\begin{align}
m_P &= \sqrt{\frac{\hbar c}{G}} &
\ell_P &= \sqrt{\frac{\hbar G}{c^3}} &
t_P &= \sqrt{\frac{\hbar G}{c^5}}
\end{align}$$

Here, since you've set three constants to have numerical values of 1, your three fundamental Planck units will be equivalent up to multiplications by factors of those three constants, $G$, $\hbar$, and $c$. In other words, multiplication by any factor of the form $G^i\hbar^jc^k$ is the equivalent to the gauge transformation I mentioned earlier. You can tell that all these units are equivalent under such a transformation, but more than that, all powers of them are equivalent! In particular, you can convert between $M$ and $M^{-1}$ by multiplying by constants whose numerical value in this unit system is equal to 1, and thus it's not a problem that $M \sim M^{-1}$ here.

# Unit systems as vector spaces

Another way of understanding unit systems, which is kind of a logical extension of the previous section, is to think of them as a vector space. Elements of this vector space correspond to dimensions of quantities, and the basis vectors can be chosen to correspond to the fundamental dimensions $L$, $T$, and $M$. (Of course you could just as well choose another basis, but this one suits my purposes.) You might represent

$$\begin{align}
L &\leftrightarrow (1,0,0) &
T &\leftrightarrow (0,1,0) &
M &\leftrightarrow (0,0,1)
\end{align}$$

Addition of vectors corresponds to multiplication of the corresponding dimensions. Derived dimensions correspond to other vectors, like

$$\begin{align}
[c] = LT^{-1} &\leftrightarrow (1,-1,0) \\
[G] = L^3M^{-1}T^{-2} &\leftrightarrow (3,-2,-1) \\
[\hbar] = L^2MT^{-1} &\leftrightarrow (2,-1,1)
\end{align}$$

In this view, setting a constant to have a numerical value of 1 corresponds to projecting the vector space onto a subspace orthogonal to the vector corresponding to that constant. For example, if you want to set $c = 1$, you project the 3D vector space on to the 2D space orthogonal to $(1,-1,0)$. Any two vectors in the original space which differ by a multiple of $(1,-1,0)$ correspond to the same point in the subspace - just like how, in the previous section, any two dimensions which could be converted into each other by multiplying by factors of $c$ could be considered equivalent. But in this view, you can actually think of the two dimensions as becoming *the same*, so that e.g. length and time are actually measured in the same unit.

Since in Planck units you set three constants to have a numerical value of one, in the dimensions-as-vector-space picture, you need to perform three projections to get to Planck units. Performing three projections on a 3D vector space leaves you with a 0D vector space - the entire space has been reduced to just a point. All the units are mapped to that one point, and are the same. So again, $M$ and $M^{-1}$ are identical, and there's no conflict.

## Best Answer

The short answer is that it is simply not possible to design a "one size fits all" unit system. The staggeringly large range of mass, time and length scales that appear in the Universe prevent this. The Planck unit system you mentioned is mainly useful for people who will never touch an experimental apparatus. The vast majority of scientists and engineers do not even do physics, let alone theoretical quantum or cosmological physics, and need a standard unit system that reflects the magnitude of quantities that are most likely to be found in their everyday work.

Thus, professional physicists use whatever unit system is most convenient for the problem at hand. There is no danger that

"more physicists grow up using the old unit system", as if that would somehow obscure people's understanding. Actually the more redundant and bizarre unit systems that trainee physicists are exposed to, the better. This teaches you fluency in converting between these systems, and helps you to communicate with people from different subfields.