It's pretty hard to answer this question when 'familiar' isn't defined further.
Every finite group (and thus every permutation group) has a composition series, which is unique in the sense that the length and composition factors of any two composition series are the same up to permutation and isomorphism. If you define familiar groups to be simple groups, then these quotients are the familiar pieces you're looking for.
Defining familiar as simple is a stretch, though. You'd be hard pressed to find someone who found the Held group especially recognizable.
The other problem is that two nonisomorphic groups can have the same composition factors. The factors do help you break the group into smaller, more recongizable chunks, but they don't tell you how those chunks interact.
There are so many variations on combinations of smaller groups that it is difficult to imagine how we could recognize them all without group presentations. Another way of breaking down a group into recognizable groups is by looking at its Sylow subgroups and seeing how they interact (this is local group theory). But for this, we need to understand $p$-groups enough to call them recognizable. My answer to this question should give you an idea of the magnitude of what we're dealing with just in the world of $p$-groups.
Take for example $\text{SmallGroup}(16,3)$. There is no other name for that group, as far as I know. It is isomorphic to $(\mathbb{Z}_4\times\mathbb{Z}_2)\rtimes \mathbb{Z}_2$, which shows us it can be split into those recognizable pieces. This decomposition uses semidirect products, however, which is basically the same as the relations / presentation concept we are trying to avoid.
So in summary, there are many a lot of different ways to divide a group up into pieces that are as familiar as possible, to gain understanding about it and how it works. However, to achieve a full description of the group, most of the time we need to use specific relations in a group presentation to show the way those pieces fit together.
Not a full answer but a suggestion that is a bit too long to fit into a comment.
The most important property about Pauli matrices is not the one you give but the following
$$ p_ip_j=\sum_k\varepsilon_{ijk}\;p_k$$
where $\varepsilon_{ijk}=\pm1$ if $(i,j,k)$ is a permutation of $(1,2,3)$. The sign $\pm1$ is the signature of the permutation. In all other cases, $\varepsilon_{ijk}=0$. It is easy to check that the property $\{p_i,p_j\}=2\delta_{ij}$ is a direct consequence of this more fundamental property.
So you should try to look for permutation matrices satisfying
$$ q_iq_jq_k=\varepsilon_{ijk\ell}q_\ell$$
and they will immediately verify $\{q_i,q_j,q_k\}=6\delta_{ijk}$.
Best Answer
For $k = 2$ the term is involutory matrix - a matrix that is its own inverse. In general matrices such that that a power of them is the identity are called matrices of finite order.
You can think about it this way - the repeated application of such a matrix to a vector must ultimately give same the vector back since repeated applications eventually lead to multiplying by the identity. That means that for a matrix where $E^k = I$ all of the eigenvalues $\lambda_i$ must satisfy $\lambda_i^k =1$. I believe that condition defines the set of matrices with this property.
See also Is permutation matrix the only matrix for which $A^k = A$