The word "cipher" in the briefing, as well as the fact that the ciphertext has been grouped into chunks of length 5, should hint that it's a simple cipher like a substitution or transposition.
The cipher used is a vigenere cipher. We can use a tool like https://guballa.de/vigenere-solver to break it knowing nothing but the ciphertext.
Alternatively, since we are given that the plaintext starts with "doubledoubletoilandtrouble..." we can decrypt the ciphertext using this as our key to get the word "christmas", which must have been our original key. (Why?)
Looking at the plaintext:
...omitted...
THISW ASAVI GENER ECIPH ERWEL LDONE ONCEA
GAINP ASTEB INDOT COMSL ASHSM ALLVB IGSBI
GMBIG KBIGX SMALL RBIGV BIGN
We can manually fix the spacing to obtain:
THIS WAS A VIGENERE CIPHER WELL DONE ONCE AGAIN
PASTEBIN DOT COM SLASH SMALL V BIG S BIG M
BIG K BIG X SMALL R BIG V BIG N
The last part looks like it tells us a URL, and indeed when we visit https://pastebin.com/vSMKXrVN we get our next ciphertext.
These values might look like decimal, but looking closely they seem a bit too large. Instead what has been used is Octal (base 8). This can be determined by using the CyberChef Magic recipe, and after decoding the result from base64 we get a plaintext consisting of the words "one" and "zero". This looks like a binary string, so let's find and replace all "zero"s with "0"s and all "one"s with "1"s and then decode the resulting binary. We obtain the flag.
Our flag is:
<aside> 💡 flag{challenge9-xAwhLJpgzpfdBnbZaTcBEXKEm}
</aside>