Bcrypt is a password hashing function likes PBKDf2, Scrypt, and, Argon, where in the password hashing the collision is not important, pre-images are important.
If you just iterate the $\operatorname{MD5^n}(x)=\operatorname{MD5}(\operatorname{MD5}(...(\operatorname{MD5}(x)...))$ n-times then we will have an already well-known problem. A collision in the inner MD5 is a collision for the $\operatorname{MD5^n}$ therefore simple iteration is not secure only slows the collision finding, really? Just find a collision for $\operatorname{MD5(x)}$ then you have a collision. In other words, the cost of finding collusion is not affected!
An easy fix is for the case $n=2$ is $\operatorname{MD5^2}(x)=\operatorname{MD5}(\operatorname{MD5}(x)||x))$ or similar approaches.
We don't need doubling MD5 or SHA-1 to improve security, we just need new hash functions like SHA3 and the very fast one Blake2b.
Finally, we want cryptographic hash functions to be secure and fast, not slow. Slowness is required in password hashing.
update for the comment
It turns out that hashing with MD5 is required for the identification. In this case, the pre-image attack is more important if the public keys are considered to be kept secret. In the pre-image attack, given a hash value $h$ we are looking for an $x'$ such that $h = \operatorname{MD5}(x')$. The $x'$ may be the original $x$ such that $h = \operatorname{MD5}(x)$ or not. If the attackers seek the original they need to search more. MD5 has only one pre-image attack that its practical cost is not faster than the generic pre-images search that has $2^{128}$-time.
The collision here is only relevant for you since you don't have two or more users who have the same identity.
So you can use a modern hash function with trimmed instead of MD5.
h=msg; for i in 1..n: h=o(h || i); endfor; return h;
? Is it still possible to reuse collisions in this case? And as I said in the comment of kelalaka's answer, I don't think reiterating the same fast hash function multiple times is the only way of creating a slow hash function. $\endgroup$