When encrypting twice with two separate keys, can a single key decrypt both steps? Is it possible that if a plaintext is encrypted first with a key, $n_1$, and then this ciphertext is again encrypted using the same algorithm but a different key, $n_2$ that the resulting ciphertext could be converted back to the original plaintext with a single key?

For example:
keys $n_1$, $n_2$, and $n_3$

$p_1$ = "plaintext"

$p_2 = C(n_1, p_1)$

$p_3 = C(n_2, p_2)$

Is  this possible:  $Dec(n_3, p_3) = p_1$?   (where $C$ is an encryption algorithm taking the key and plaintext)

An example I can think of is the Caesar Cipher (the key can be seen as the shift). Where first a shift of 4 is used, and then a shift of 6. A shift of 16 would reveal the plaintext in a single operation (a single key).

Obviously this is a trivial example. Are there any cases where this could happen with a stronger algorithm, such as AES?
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.