Hypothetical unknown cipher - security in obscurity? I'm curious what would happen in the following scenario:

Suppose an attacker gets a hold of a cipher-text of sufficiently large length. And suppose he has the means to verify a successful decryption.
But the attacker has no clue what cipher is used and assumes it's unbreakable with anything but brute-force whatever it is.
He also assumes it's a block cipher. (Thus, whatever operations the cipher performs are repeated over each block.)


Suppose he knows the block is of length 128bit. How long (what is the complexity) will it take him to brute-force all possible operations that can be performed?
What if it's XTS and the attacker doesn't know anything about any standards?


I guess this is just to contrast the security in obscurity argument, but I'm also interested in the complexity numbers.

Also, what would happen if you modify AES in some way, like adding some extra rounds or an extra operation, something simple. And make it such that the change is NOT hard-coded into the algorithm but is instead loaded as a module/script as PART of the key to access the encrypted data? How much complexity would that add to someone trying to get your plain-text if he doesn't even know precisely what algorithm is used.
An unknown block-cipher with an $n$-bit block length can just be viewed as an arbitrary permutation of size $2^n$. Assuming you have the means to verify a correct decryption, you can just try each of every $2^n!$ possible permutations.

Naturally, if your ciphertext consists of fewer than $2^n$ distinct block values then you can omit permutations which result in previously-tried plaintexts.

An extreme example of this would be a ciphertext consisting of one block of $n$ bits which would need only a maximum of $2^n-1$ decryptions to be tested. This is of course significantly less than $2^n!$