First, let's derive the 6 sets. Note that order doesn't matter so instead of getting 6 sets we'll have 6x5! = 720 sets.
We need to find 5 gamma products that anticommute. There are 16 products available but we can't use 1 so there are 15 choices. Without loss of generality (because the gamma matrices are equivalent, up to multiplication by $\pm 1, \pm i$ to change the square from +1 to -1 or back), we can assume we chose $\gamma^0$
The second choice needs to anticommute with our first choice $\gamma^0$. There are 8 gamma products that anticommute and 8 that commute. (Of the 8 that commute, one is our choice and another is 1.) For the first choice of $\gamma^0$, the eight we have available to choose for the second anticommuting matrix is given by the eight products in $\{1,\gamma^0\} \;\times\; \{\gamma^1,\gamma^2,\gamma^3,\gamma^1\gamma^2\gamma^3\}$, i.e. choose one of the first two (that commute with $\gamma^0$) and multiply it by one of the last four (that anticommute) so the product will anticommute. The eight cases give the 8 gamma products that anticommute with $\gamma^0$. Our choice doesn't matter, we will make the canonical choice of $\gamma^1$.
The third choice is one of those same 8 but now we can't use $\gamma^1$ and we also can't use the product of our first two choices $\gamma^1\gamma^0$. Looking through the 8 possibilities, there are 3 that anticommute with the first two choices and are also not a product of them. They are $\{\gamma^2,\gamma^3,\gamma^0\gamma^1\gamma^2\gamma^3\}$. Our choice doesn't matter, we will make the canonical choice of $\gamma^2$.
The fourth choice has to be one of the remaining two possibilities $(\gamma^3,\gamma^0\gamma^1\gamma^2\gamma^3)$ the canonical choice is $\gamma^3$ and that defines the fifth choice as the product of the other four.
The number of choices for these 4 stages was 15,8,3,2 and the product of these is 720 so the number of ways of defining sets of 5 anticommuting gamma products. Taking into account ordering, we have 720/5! = 720/120 = 6 cases which are listed in the reference.
Now the OP question was "what is the relation between these 6 sets?" Since each of these sets has an arbitrary ordering, it might be more productive to ask instead, "what is the relation between these 720 ordered sets?"
For standard physics, these 720 sets of 5 anticommuting matrices are all equivalent, so it doesn't matter which we choose. However, if we make the assumption that the gamma matrices are a part of physical reality, (and are not just an arbitrary mathematical choice used to obtain a symmetry compatible with experimental observation), it's possible that we can discover something that is a consequence of our choice of which 4 we choose as $\gamma^0,\gamma^1,\gamma^2,\gamma^3$. For example, imagine physics where different elementary particles are modeled with different gamma matrices.
If we are going to do that sort of thing, then it might be also useful to note that we can multiply gamma products by $-1$ without making much of a change to them. Proper rotations cannot introduce a single minus sign so if we want to add minus signs we need to bring them in 0, 2 or 4 at a time. The number of ways of doing that are 1, 10 and 5 which will increase the number of gamma matrix sets by a further factor of 1+10+5 = 16.
A simpler problem is to consider these sorts of transformations on the Pauli spin matrices. For these, it turns out that there are 24 choices for proper rotations and they make up the point symmetry 432 which the crystallographers call gyroidal. That finite group has 24 elements and is the same as the permutation group for 4 elements called $S_4$. It is a finite subgroup of the spin-1 rep of SU(2) or SO(3). You will find $S_4$ used in the published phenomenology literature quite a lot (but not in reference to gamma matrix symmetries discussed here), for example https://doi.org/10.1016/j.nuclphysb.2018.12.016
In looking for a transformation from one of the six sets to another, we're working in 4x4 matrices so the transformation is going to use unitary 4x4 matrices. I've indicated that with the Pauli matrices these sorts of things fit into SU(2) so I expect the gamma matrix calculations to be SU(4) calculations and indeed that happens, and since SU(4) is a subgroup of U(4) the transformations are indeed unitary.
The first two example lines of the reference are:
$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5$ and $y_1,y_2,y_3,y_4,y_5$. We only need to do the first four of each as the fifth is just the product of the first four. These are defined in terms of $\sigma_j$ and $\rho_j$ as:
\begin{array}{rclrcl}
\alpha_1 &=& \rho_1\sigma_1,&y_1&=&\rho_2\sigma_1\\
\alpha_2 &=& \rho_1\sigma_2,&y_2&=&\rho_2\sigma_2\\
\alpha_3 &=& \rho_1\sigma_3,&y_3&=&\rho_2\sigma_3\\
\alpha_4 &=& \rho_3\sigma_0,&y_4&=&\rho_3\sigma_0
\end{array}
so the transformation we want needs to take $\rho_1$ to $\rho_2$, and it needs to leave the $\sigma_j$ and $\rho_3$ alone. Since the $\rho_j$ and $\sigma_k$ commute, this means our transformation will depend on $\rho_k$ alone. And since we want to leave $\rho_3$ alone, our basis set should commute with $\rho_3$. That leaves only two elements in the basis set of the transformation: $\{1,\rho_1\rho_2\}$. Note that $\rho_1\rho_2 = i\rho_3$ but I'll leave it as $\rho_1\rho_2$ cause I think it makes the transformation more obvious.
Let $U(\theta) = \exp(\theta\rho_1\rho_2)$ be a unitary matrix. Here we are writing the unitary matrix as $\exp(iH)$ where $H$ is an Hermitian matrix. Check that indeed, $i^2\rho_3$ is Hermitian. Yep. Now compute.
\begin{equation}
\exp(\theta\rho_1\rho_2) = 1 + \theta\rho_1\rho_2 + \theta^2(-1)/2! + \theta^3(-\rho_1\rho_2)/3! ...
\end{equation}
where we simplify using $(\rho_1\rho_2)^2=-1$ by anticommutativity of $\rho_1$ and $\rho_2$ and this simplifies to
\begin{equation}
U(\theta) = \cos(\theta) +\rho_1\rho_2\sin(\theta)
\end{equation}
It's clear that $U(\theta)$ commutes with $\rho_3$ so let's see what it does to $\rho_1$. Here we will use commutation rules between $\rho_1$ and $\rho_2$ to move the exponential around the object being transformed ($\rho_1$). This transformation negates the exponential:
\begin{equation}
\begin{array}{rcl}
U(-\theta)\;\rho_1\;U(\theta) &=&
\exp(-\theta\rho_1\rho_2)\;\rho_1\;\exp(\theta\rho_1\rho_2),\\
&=&\rho_1\;\exp(+\theta\rho_1\rho_2)\exp(\theta\rho_1\rho_2),\\
&=&\rho_1\;\exp(2\theta\rho_1\rho_2),\\
&=&\rho_1(\cos(2\theta)+\sin(2\theta)\rho_1\rho_2),\\
&=&\cos(2\theta)\rho_1+\sin(2\theta)\rho_2.
\end{array}
\end{equation}
Now to get this equal to the desired $\rho_2$ we need $\sin(2\theta)=1$ and $\cos(2\theta)=0$ So $2\theta = \pi/2$ and $\theta = \pi/4$ and the desired $U$ is
\begin{equation}
\begin{array}{rcl}
U(\pi/4) &=& \exp((\pi4)\rho_1\rho_2) = (1+\rho_1\rho_2)/\sqrt{2},\\
U(-\pi/4) &=&(1-\rho_1\rho_2)/\sqrt{2}
\end{array}
\end{equation}
And the reader is invited to verify that this $U$ transforms $\rho_1$ to $\rho_2$. In fact, thinking of the $\rho$ as Pauli spin matrices, what we've done is derived the unitary transformation corresponding to a rotation by 90 degrees around the z axis. This leaves $\rho_3$ alone and takes $\rho_1$ to $\rho_2$, as desired.
