When encrypting twice with two separate keys, can a single key decrypt both steps?
It is possible with the old Pohlig-Hellman cipher.

Here's how it works:


The global parameter is a prime $p$
A secret key is a value $k$ which is relatively prime to $p-1$
To encrypt a message $M$, you compute $C = M^k \bmod p$, and that's you
To decrypt a ciphertext $C$, you compute $k^{-1} \bmod p-1$, and then compute $M = C^{k^{-1}} \bmod p-1$


With this system, if you encrypt with $k_1$, and then with $k_2$, the resulting ciphertext is $C = (M^{k_1} \bmod p)^{k_2} \bmod p = M^{k_1k_2} \bmod p$; hence, this is identical to encrypting with the key $k_1k_2 \bmod p-1$, and so can be decrypted by that key.