What was the key length of the Enigma? When I type in any plaintext to an Enigma machine (Enigma machine emulator) the substitution does not seem to follow any pattern. In other words, there is no cycle in the use of enciphering alphabets (defined as the number of switches done when turning a plaintext letter to a ciphertext letter - which can vary from 0 to 26). 

But since the key is the alignment of the rotors, there must be a finite key length even though it probably is longer than any plaintext ever would be. I imagine if you keep writing long enough, the rotors will eventually align in a way they already were. If I'm right - how long is the key?

And if the key is much longer than any plaintext, wouldn't that mean the Enimga is a one-time pad? From what I can read, the OTP is defined as an algorithm with a key longer than the plaintext. Is this wrong?
Enigma works with the idea of one time pad, it's a machine that automates key generation. In the one time pad, we have to generate a random key with the same length as plaintext but in Engima we make few setups, and the machine creates a that random key.
Here is the enigma procedure

A day key has the form


Plugboard setting: A/L–P/R–T/D–B/W–K/F–O/Y
Scrambler arrangement: 2-3-1
Scrambler starting position: Q-C-W


Sender and receiver set up the machine the same way for each message
Use of message key: a new scrambler starting
position, e.g., PGH


first, encrypt and send the message key, then set the
machine to the new position and encrypt the message
initially, the message key is encrypted twice