Question about Enigmas involutional property and its reduction of possible ciphers
The number of involutions of size $n$ satisifies the recurrence relation:

$$I(n) = I(n-1) + (n-2)I(n-2)$$

(This recurrence is easy to deduce).

Given than $I(1) = 1$ and $I(2) = 2$, we can easily compute $I(4) = 8$, which is strictly smaller than the number of permutations of size 4 (which is 24).

This recurrence allows us to quickly compute the number of involutions of size 26 as 158,432,124,870,784; much smaller than the number of permutations 26! = 604,937,191,689,908,372,404,961,280

BTW: the involutions of size 4 are:

$$ABCD$$

$$ABDC$$

$$ACBD$$

$$ADCB$$

$$BACD$$

$$BADC$$

$$CBAD$$

$$CDAB$$

$$DBCA$$

$$DCBA$$