The permutation matrices for $2$ and $3$ dimensions look like this:
- $2$-dimensional:
$$ M_1^{2d} = \left(\begin{matrix} 1 &0\\ 0 &1\end{matrix}\right), \qquad M_2^{2d} = \left(\begin{matrix}0 &1\\1 &0\end{matrix}\right) $$
- $3$-dimensional:
$$ M_1^{3d} = \left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right), \qquad M_2^{3d} = \left(\begin{matrix}1&0&0\\0&0&1\\0&1&0\end{matrix}\right), \qquad M_3^{3d}=\left(\begin{matrix}0&1&0\\1&0&0\\0&0&1\end{matrix}\right)$$
$$ M_4^{3d} = \left(\begin{matrix}0&0&1\\0&1&0\\1&0&0\end{matrix}\right), \qquad M_5^{3d} = \left(\begin{matrix}0&1&0\\0&0&1\\1&0&0\end{matrix}\right), \qquad M_6^{3d} = \left(\begin{matrix}0&0&1\\1&0&0\\0&1&0\end{matrix}\right)$$
Matrices $M_1^{2d}$ and $M_1^{3d}$ are the identity matrices. Furthermore, $M_2^{3d}, M_3^{3d}, M_4^{3d}$ are permutations of only two elements (thus their action is described by only $M_2^{2d}$).
The only complete permuations (where each of the elements is permutated) are $M_2^{2d}$, $M_5^{3d}$, $M_6^{3d}$. One conditions for that is $\mbox{Tr}(M)=0$.
Does this subset of permutation matrices have a special name? Is there a way to construct them easily?
It looks like you want the permutations corresponding to all cycles of length $d$. We can characterize these permutations as follows:
Let $\sigma_d$ denote the permutation $i \mapsto i+1 \pmod d$, corresponding to the matrix $$ \pmatrix{ &&&&1\\ 1\\ &1\\ &&\ddots\\ &&&1 } $$ The cycles of length $d$ on $d$ elements are exactly those that can be written as $\tau \sigma _d \tau^{-1} $, where $\tau$ is an arbitrary permutation on $d$ elements.
In general, there will be $(d-1)!$ such permutations out (of the $d!$ total permutations).
Such a permutation that is not simply a power of $\sigma_d$: take $1 \mapsto 2 \mapsto 4 \mapsto 3$, as given by $$ \pmatrix{ &&1\\ 1\\ &&&1\\ &1 } $$ The above is not a power of $$ \pmatrix{ &&&&1\\ 1\\ &1\\ &&1 } $$