There are 24 different letters in the first cipher text and 24 different letter in the second one.

There are 42504 (24!/((24-5)! 5!)) possibilities to choose 5 letters from set of 24 different letters.

So this problem is solvable by the most basic brute-force checking all 42504 possibilities.

Each possibility is represented by a ciphertext from which 5 different letters has been removed.

(each 5-letter set represents possible guess for camouflage letters of which we get rid off)

We do not have to solve both ciphertexts, because if we found solution for the first one, then using same key for decryption we can decipher also the second one.

So there are 42504 possibilities to check for possible monoalphabetic substitution.

I think it is possible to solve it by brute-force.

Another way to solve it:

The camouflage alphabet is very short compared to the plaintext alphabet. The shorter the camouflage alphabet is the bigger number of occurrence there is for particular letter of camouflage.

For example if camouflage alphabet was only one letter long, say "a", than to camouflage whole text we need to use it many times, which increase its occurrence it the ciphertext, so it means that the letter with the most frequency of occurrence would be with high probability the camouflage one.

Here we have camouflage alphabet of length 5, which means that there is high probability that letters that represents camouflage are among the more frequent letters.

Other way said, letters in ciphertext with small frequency are most likely the meaningful one (i.e. not camouflage). But there is still chance that, say ,one letter of camouflage can occur among the letters of small frequency.

So we count occurrence of letters of both ciphertexts **separately**, then choose upper bound for occurrence, say 3, and we presume that all characters that occur less or equal to 3 are the meaningful one.

Other way said, we presume that all camouflage letter are among letters of occurrence 4 or more.

So we have two sets… one set represents most frequent letters from first cipher text and second one from second ciphertext.

We know that both ciphertext contain same letters of camouflage, so we make an intersection of the mentioned two sets. This intersection reduces the possibilities of letters from we have to choose 5 letters that represents camouflage.

So now we do not have to check all 42504 cases, but we reduced it to number slightly bigger than 100 possibilities to check. (we do not have to choose 5 from 24, but 5 from much smaller number)

Now we can use brute-force attack only on those reduced set.

But we have additional information that there is a word of same spelling in both ciphertexts and we know that plaintexts have same meaning so it is probable that there are words in English which have their equivalents in German. All this additional should help you to find the solution even without a brute-force.

Hope it is understandable. My Explanation in English is limited.

If this was not enough then I will disclose one letter of camouflage, but I hope this would not be necessary.