Lehmer random number generator - cipher? Let's consider linear congruential generator:
$X_{k+1} = a \cdot X_{k} \mod 2^{128}$
Such that $a$ is some number which for every 128-bit input $X_{k}$ from $0$ to $2^{128}-1$ will give us different output $X_{k+1}$ from $0$ to $2^{128}-1$. So we got bijection here (we can find many such $a$).
Now let's say we will choose such 128-bit $a_{1},a_{2}, ..., a_{10}$ as a keys, randomly. We make $10$ rounds of encryption and every round compute:

$a_{i} \cdot INPUT \mod 2^{128}$
Reverse 128-bit block.

So the next round input is an inverted block of previous round output. And it looks like it make a big difference. If we wouldn't reverse block - we got just keyed Lehmer random number generator with a modulus which is a power of two - and it is definitely not secure (because the low $k$ bits form a modulo-$2^{k}$ generator all by themselves).
But with &quot;reversing&quot; we get completely different results! It looks chaotic. I have no idea is 10 rounds enough - just guessing. Is this type of encryption secure? Do you see any obvious weaknesses? To analyze it we can choose smaller keys and smaller operating mode - for example 5-bit or 10-bit. A detailed analysis of this cipher can be difficult, my question is rather about some obvious weaknesses. The cipher appears to be unexpectedly good, despite being only a minor modification of the linear congruential generator. So either I don't notice something or it is quite a non-trivial proposition.
Comment 1: let's skip the encryption problems with zero-block - it can be solve easily, for example if we will use xoring before every round.
Is this type of encryption secure?

Nope (even if you make all the $a_i$ values odd, which you need to make it invertible).
It turns out that there are high probability differentials in this structure; in particular, you have a one-round differential with an input xor difference of (01111...11110) (that is, all bits except the first and last are flipped) and the output xor difference of (01111...11110); this differential holds with probability $&gt; 0.5$.  Because the output differential of round $X$ is the input differential of round $X+1$, the entire differential holds through the entire cipher with probability $&gt; 2^{-10}$