Here's the conversion. I will write Gaussian quantities as
$$\hat{φ}, , , , , , , \hat{ρ}, q, \hat{ε}, \hat{μ}$$
and the corresponding SI quantities as
$$φ, , , , , , , ρ, e, ε, μ.$$
Kinematic and mechanical quantities $$ (velocity), $$ (momentum), $m$ (rest mass), $M$ (moving mass), $E$ (energy), $$ (force), $P$ (power) do not differ between the two. I will also use $$ to denote the Maxwell-Lorentz Lagrangian density, which is also the same for the two.
Corresponding to the equations in Gaussian form:
$$
= ∇×, \hspace 1 em = -∇\hat{φ} - \frac{1}{c} \frac{∂}{∂t}, \hspace 1em ∇· = 0, \hspace 1em ∇× + \frac{1}{c}\frac{∂}{∂t} = , \\
∇· = 4π\hat{ρ}, \hspace 1em ∇× - \frac{1}{c}\frac{∂}{∂t} = \frac{4π}{c}, \hspace 1em ∇· + \frac{∂\hat{ρ}}{∂t} = 0, \\
= \frac{||^2 - ||^2}{8π}, \hspace 1em = \hat{ε}, \hspace 1em = \hat{μ}, \\
= q\left( + \frac{×}{c}\right), \hspace 1em P = q·
$$
are the equations in SI form:
$$
= ∇×, \hspace 1 em = -∇φ - \frac{∂}{∂t}, \hspace 1em ∇· = 0, \hspace 1em ∇× + \frac{∂}{∂t} = , \\
∇· = ρ, \hspace 1em ∇× - \frac{∂}{∂t} = , \hspace 1em ∇· + \frac{∂ρ}{∂t} = 0, \\
= \frac{ε_0||^2}{2} - \frac{||^2}{2μ_0}, \hspace 1em = ε, \hspace 1em = μ, \\
= e\left( + ×\right), \hspace 1em P = e·,
$$
with $c = 1/\sqrt{ε_0μ_0}$.
The conversions are:
$$
(\hat{φ}, , , ) = \sqrt{4πε_0} (φ, c, c, ), \\
(, , , \hat{ρ}, q) = \frac{1}{\sqrt{4πε_0}}\left(4π, \frac{4π}{c}, , ρ, e\right), \\
(\hat{ε}, \hat{μ}) = \left(\frac{ε}{ε_0}, \frac{μ}{μ_0}\right).$$
In the reverse direction, they are
$$
(φ, , , ) = \frac{1}{\sqrt{4πε_0}}\left(\hat{φ}, \frac{}{c}, \frac{}{c}, \right), \\
(, , , ρ, e) = \sqrt{4πε_0}\left(\frac{}{4π}, \frac{c}{4π}, , \hat{ρ}, q\right), \\
(ε, μ) = (ε_0 \hat{ε}, μ_0 \hat{μ}).
$$
There is no direct mention of $ε_0$ or $μ_0$ in the Gaussian version, only of $c$; while in the SI version, there is no direct mention of $c$, only of $ε_0$ and $μ_0$ and it's confined to the (vacuum version of the) constitutive relations and Lagrangian density, nowhere else. All the other equations are $c$-independent and (in fact) non-metrical. They live at a deeper level of geometry where there is no concept of a metric, parallelism, orthogonality, congruence, speed, or any distinction between space-like and time-like.
If the dimensions are denoted $Q$, $P$, $M$, $L$ and $T$, respectively for electric charge/flux, magnetic charge/flux, mass, length and time duration, then the dimensions for the various quantities in SI will be
$$[\left(φ, , , , , , , ρ, e, ε, μ\right)] = \left(\frac{P}{T}, \frac{P}{L}, \frac{P}{L^2}, \frac{P}{LT}, \frac{Q}{L^2}, \frac{Q}{LT}, \frac{Q}{L^2T}, \frac{Q}{L^3}, Q, \frac{QT}{PL}, \frac{PT}{QL}\right).$$
The dimensions of $P$ and $Q$ are conjugate and multiply out to those of action:
$$PQ = \frac{ML^2}{T}.$$
From this, you can determine what the dimensions for the Gaussian quantities are:
$$[\left(\hat{φ}, , , , , , , \hat{ρ}, q, \hat{ε}, \hat{μ}\right)] = \left(\frac{G}{L}, \frac{G}{L}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2T}, \frac{G}{L^3}, G, 1, 1\right), \hspace 1em G = \sqrt{\frac{ML^3}{T^2}}.$$
They are, needless to say, awkward - square roots of dimensions involving $M$, $L$ and $T$. But at least they got rid of the $P$'s and $Q$'s. You might say that they paid no mind to the $P$'s and $Q$'s.
Finally, the kinematic and mechanical dimensions are independent of the system:
$$[(, P, E, , m, M, c, , )] = \left(\frac{ML}{T^2}, \frac{ML^2}{T^3}, \frac{ML^2}{T^2}, \frac{ML}{T}, M, M, \frac{L}{T}, \frac{L}{T}, \frac{M}{LT^2}\right).
$$
