This isn't a complete answer, and it partly reproduces, from a different angle, what Ilmari has already written, but I'll show how questions about *linear* invertible rules for the Margolus neighborhood can be translated into calculations in a nice commutative group algebra over the cell-state space $\mathbb{F}_2$. (I'm not sure whether this might help to complete Ilmari's argument.)

Linearity over $\mathbb{F}_2$ reduces to additivity, and short-period phenomena are going to arise when all translations of the grid state have order a power of two, resulting in massive cancellation of even numbers of terms - ultimately the same mechanism that causes the even-odd patterns in Pascal's triangle.

The details are fiddly, and far from uniform, even though there aren't that many possible rules. The period can be unexpectedly short; the half-period sometimes but not always transports the starting pattern to the antipode (halfway around the square grid torus in any diagonal direction) and sometimes there's a wallpaper of other lit cells besides, or patterns may travel vertically or horizontally rather than diagonally, and sometimes the evolution doesn't even reach the antipodal points. I'll include some examples in this answer and I'm going to add a few more in a separate community-wiki answer.

### Setup

A single iterate for the Margolus neighborhood consists of *two* steps as the web simulator counts them: One application of the rule - the odd-numbered step - to each $2\times 2$ block on the grid as given (the lighter grey grid markings are on the midpoints of the blocks), followed by another application to each $2\times 2$ block of the "dual" grid (now with the block corners at the grid markings) - the even-numbered step. (For simplicity I'll limit myself to using the same per-block rule in odd and even steps.)

Also, it will be convenient to regard the grid as consisting of $M\times M$ blocks, rather than $2M\times 2M$ cells as the simulator's notation would suggest. A state of the grid is thus a mapping $\mathbb{Z}/m \times \mathbb{Z}/m \rightarrow (\mathbb{Z}/2 \times \mathbb{Z}/2 \rightarrow \mathbb{F}_2)$ from the grid-of-blocks to block states.

Thus OP's original example has period 224 (not 448) on a grid of $32\times 32$ blocks, lifted from period 7 on a single-block grid.

The simulator uses numbers $\{0,\dotsc,15\}$ to denote block states, where $1,2,4,8$ respectively indicates that the northwest, northeast, southwest, southeast cell is lit (nonzero), with bitwise addition. Rules are entered by listing the image of each state in order; for invertible rules, this will be a permutation of $\{0,\dotsc,15\}$. Linear rules of course must leave the state $0$ fixed, and are determined by the images of the basis states $1,2,4,8$ by additivity. A priori they correspond to elements of $\DeclareMathOperator{\GL}{GL} \GL_4(\mathbb{F}_2)$. Eliminating conjugates under reflections and rotations of the grid, however, leaves just 2606 distinct possibilities (if I got this right), and eliminating inverses of (conjugates of) earlier rules as well reduces this list to only 1423.

### Linear maps acting on grid states

We'll write $R$ for the given rule, considered as acting during an odd step on each block of the grid simultaneously.

One of the simulator's predefined rules is linear and invertible: "Swap On Diag", or `0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15`

, or $1\leftrightarrow 8, 2\leftrightarrow 4$. This calls for the letter $X$ to denote the induced map on grid states.

Now we can already write down how a single iterate $A$ acts on a single-block grid: Namely, as $XRXR$ - an odd step $R$ followed by an even step $XRX$.

Note that while $XRXR$ is still a linear invertible map of block states, $R\mapsto XRXR$ isn't an endomorphism of $\GL_4(\mathbb{F}_2)$. It turns the 1423 candidate rules into just 1110 distinct permutations of block states, with just eight possible cycle structures: one 15-cycle, or two 7-cycles (as in OP's example), or three 5-cycles, or four or five 3-cycles, or four or six transpositions, or the identity. Let $p$ be the order of this permutation.

Returning to square grids of more than one block, let's write $\newcommand{\N}{\mathrm{N}} \newcommand{\W}{\mathrm{W}} \newcommand{\E}{\mathrm{E}} \newcommand{\NW}{\mathrm{NW}} \newcommand{\NE}{\mathrm{NE}} \newcommand{\SW}{\mathrm{SW}} \newcommand{\SE}{\mathrm{SE}} T_{\NW}, T_{\N}, T_{\NE},\dotsc$ for the translations that shift a pattern by one block's worth to the northwest, north, northeast, and so forth. On a torus grid of $M\times M$ blocks, these generate an elementary abelian group $T$ isomorphic to $\mathbb{Z}/M \times \mathbb{Z}/M$. For convenience, let $T_0$ be the identity translation. The translations commute with all the other linear maps on grid states we're considering.

For any $M, m \in \mathbb{N}$, there's a contraction (or folding) map $C_{Mm,m}$ from $Mm \times Mm$-grid states to $m\times m$-grid states, by subdividing the larger grid into $M\times M$ tiles and summing the tile states. Strictly speaking these maps depend on choices of origin, but since translations commute with everything else in sight, this doesn't matter. And the contractions also commute with everything else. So by considering $C_{M,1}$, the period of the iterated rule application on an $M\times M$ grid must be a multiple of $p$.

Finally, there are orthogonal idempotents $P_{\NW}, P_{\NE}, P_{\SW}, P_{SE}$ on the space of grid states which respectively leave the northwest, northeast, southwest, or southeast cells in all blocks unchanged and turn off all the other lights.

We can use them to algebraize what $R$ does during a single step: E.g. if $P_{\NW}RP_{\SE}$ is a nonzero map, then $R$ during the odd step will light a NW cell given a lit SE cell (absent other contributions). Think of this if you like as extracting "matrix elements" from $R$.

These projections *don't* commute with everything else, but they play nicely with $X$, as in $P_{\NW}X = XP_{\SE}$ etc.

### Additive decomposition of the rule application

Now we can write down what the even steps do. Thanks to linearity, we only need to work out the summands arising from a single lit cell in the input state and contributing to a single cell's output state. For example, the SE cell of the next block to the NW receives a contribution only from a lit NW cell of the origin block, and only if the rule has a suitable nonzero matrix element:
\begin{gather}P_{\SE}T_{\NW}XRXP_{\NW} \\
= P_{\SE}T_{\NW}X(P_{\NW}RP_{\SE})XP_{\NW}\end{gather}

The same input cell could also contribute to the SW cell's state in the next block to the north:
\begin{equation}P_{\SW}T_{\N}XP_{\NE}RP_{\SE}XP_{\NW}\end{equation}
as well as to its own future state:
\begin{equation}P_{\NW}T_0XP_{\SE}RP_{\SE}XP_{\NW}\end{equation}

And so on.

The linear map on grid states executing an even step is just the sum of all expressions of these forms. And the linear map $A$ representing a single iterate (odd step followed by even step) is obtained by multiplying them, from the right, with the odd-step "matrix elements" like $P_{\NW}RP_{\SE}$ in all possible combinations. Only those products where adjacent projections match can yield nonzero contributions.

Patterns racing at light speed to the northwest, for example, come from (lit SE cells in the initial state and) a single summand:
\begin{equation}P_{\SE}T_{\NW}XP_{\NW}RP_{\SE}XP_{\NW}RP_{\SE}\end{equation}
and we can "see" them move in the group algebra $\mathbb{F}_2[T]$ by noting that when we compose the iterates, we can rearrange all the $T_{\NW}$ to the left of the product and put them next to each other.

Example 1

The Swap On Diag rule $R=X$ consists only of the four summands of the above type - all SE cell excitations glide NW, all SW ones glide NE, etc., and they all return to the point of origin after $M$ iterations, and (as long as $M$ is even) combine to form a clean antipodal image of the input pattern after $M/2$ iterations. Thus $p=1$ (since $X$ is an involution, thus $XRXR$ is the identity) and the overall period equals the grid diameter $M$ (illustrated below for $M=32$; the animation consists of 33 frames, repeating the starting state to emphasize the period).

Example 2

Similarly, the rule `0,2,1,3,8,10,9,11,4,6,5,7,12,14,13,15`

or $1\leftrightarrow 2, 4\leftrightarrow 8$ sends all SE and NE cells at light speed to the west (summands with $T_{\W}$), and vice versa, again giving $p=1$ and period $M$.

Example 3

In the variant `0,1,2,3,8,9,10,11,4,5,6,7,12,13,14,15`

or $1\leftrightarrow 2, 4\mapsto 4, 8\mapsto 8$, on the other hand, the gliders never go far: Now $p=2$ and the overall period remains 2 regardless of grid size, showing that any inductive argument exploiting the contraction maps $C_{2M,M}$ must start from a $2\times 2$-blocks grid as the base case, not from the single-block case! Spelling it out in the above formalism, there are terms with $T_{\W}$ and with $T_{\E}$, but the only products with matching projections in the second iterate end up with $T_{\W}T_{\E} = T_{\W}T_{\E} = T_0$ as the translation.

The second and third example do not depend on $M$ being a power of 2.

### A final simplification

Fortunately, it turns out to be unnecessary to keep track of all the iterated $XPRPXRP$ business. Once we've computed the single iterate $A$, we can simply suppress all terms that contain vanishing matrix elements $PRP$, and all products where two different projection idempotents meet each other, and keep the surviving summands, with the mess hidden inside a black box: E.g., what the $k$-th iterate of the northwest-gliding summand does (if $P_{\NW}RP_{\SE}$ is nonzero) is fully captured by
\begin{equation}P_{\SE}T_{\NW}^k\blacksquare P_{\SE}\,.\end{equation}

Example 4

Among the simplest nontrivial rules is `0,1,4,5,2,3,6,7,9,8,13,12,11,10,15,14`

or $1\mapsto 1, 2\leftrightarrow 4, 8\mapsto 9$. The lit SW cells race NE and vice versa, independently of anything else, so I'll be omitting them from the following expressions. The NW and SE cells couple to each other; their contribution to the single iterate $A$ is:
\begin{align}
P_{\NW} (T_0) \blacksquare P_{\NW} & + \\
P_{\NW} (T_0) \blacksquare P_{\SE} & + \\
P_{\SE} (T_{\NW}) \blacksquare P_{\NW} & + \\
P_{\SE} (T_{\NW} + T_0) \blacksquare P_{\SE}\end{align}

Squaring this expression and collecting the translations gives:
\begin{align}
P_{\NW} (T_{\NW} + T_0) \blacksquare P_{\NW} & + \\
P_{\NW} (T_{\NW}) \blacksquare P_{\SE} & + \\
P_{\SE} (T_{\NW}^2) \blacksquare P_{\NW} & + \\
P_{\SE} (T_{\NW}^2 + T_{\NW} + T_0) \blacksquare P_{\SE}\end{align}

And the third iterate is:
\begin{align}
P_{\NW} (T_{\NW}^2 + T_{\NW} + T_0) \blacksquare P_{NW} & + \\
P_{\NW} (T_{\NW}^2 + T_0) \blacksquare P_{SE} & + \\
P_{\SE} (T_{\NW}^3 + T_{\NW}) \blacksquare P_{NW} & + \\
P_{\SE} (T_{\NW}^3 + T_{\NW}^2 + T_0) \blacksquare P_{\SE}\end{align}

Contracting this to a $1\times1$ grid by setting all translations equal to $T_0$, we get the identity map (NB also on the omitted SW and NE terms), thus $p=3$.

Contracting it to a $2\times 2$ grid amounts to setting the squares of all translations equal to $T_0$; the middle two terms vanish due to cancellation modulo 2, and we're left with the antipodal translation (again, also on the SW and NE terms which I've omitted):
\begin{align}
P_{\NW} (T_{\NW}) \blacksquare P_{\NW} & + \\
P_{\NW} (0) \blacksquare P_{\SE} & + \\
P_{\SE} (0) \blacksquare P_{\NW} & + \\
P_{\SE} (T_{\NW}) \blacksquare P_{\SE}\end{align}
which is an involution on the space of grid states, confirming the overall period 6.

Lifting this inductively to $M\times M$ grids where $M$ is a power of 2 now reduces to a computation with binomial and multinomial coefficients modulo 2.

### In general...

Further complications come from translation relations like $T_{\NW} = T_{\W}T_{\N}$.

We're left having to perform computations in $\mathbb{F}_2[T]$. The summands could be organized by paths in a Cayley graph of the group of translations $T$, with individual steps depending on the nonzero matrix elements of $R$, but I'm not sure whether that helps.

### Afterthought:

We don't even need the black boxes! All these $P_{\NW} (T_{\NW}^2 + T_0) \blacksquare P_{\SE}$ (etc.) terms *are* bona fide matrix elements - and we need to keep only their $T_{\NW}^2 + T_0$ (etc.) parts when we arrange them into a $4\times 4$ matrix (which we may again call $A$) whose rows and columns are indexed by $\{\NW,\NE,\SW,\SE\}$. So we can work in a matrix algebra over the commutative algebra $\mathbb{F}_2[T]$, and in fact in the commutative matrix subalgebra generated by $A$.

But I still don't see how to extract from this a uniform inductive argument that would correctly predict the periods for *all* linear invertible rules on grids of power-of-two diameter.

muchlonger periods. Your example does seem to hit a special case. It is even more special: after ahalfperiod, it replicates the input pattern but translated to the antipode on the torus (e.g. shifted by (32,32) on a 64x64 grid). Is this rule perhaps invariant under some kind of scaling transformation - some discrete analogue of inflation/deflation...? $\endgroup$ – GNiklasch Oct 12 '17 at 13:58linearityis special (most of the predefined rules aren't: "Swap on Diag" is, and is one of the simplest examples, and visually boring). Second, when you have a linear rule, you can take cell coordinates mod 2 and look at the induced action (with the same rule!) on a 2x2 grid. And then decompose the upstairs action into a part lifted from downstairs and a translation-by-blocks part. Your $p$ will come from the induced permutation of the 16 states of the 2x2 grid (aftertwosteps: one for each block position/phase). $\endgroup$ – GNiklasch Oct 13 '17 at 17:21