The probability that a prime p does not divide a random integer n is (1-1/p), so for random n we could argue that the probability that n and φ(n) are coprime is

$\prod_{p|n} \left(1-1/p \right) = \phi(n)/n.$

The average order of φ(n)/n is given by

${ 1 \over N } \sum_{n=1}^N {\phi(n) / n} = 6/\pi^2 + O(\log N/N).$

Now the probability that a random integer is squarefree is $6/\pi^2$.

So my question is: does gcd(n,φ(n))=1 for almost all squarefree n? Or to put it another way, for random squarefree n is the probability that n and φ(n) are coprime one? (Of course we have gcd(55,φ(55))=5, etc.)

I have not been able to find anything about this on the internet and so would like to know if this has been considered before. Thanks.

EDIT: Take integer N and let f(N) = number of squarefree n<=N such that gcd(n,φ(n))>1 (e.g. 21 or 55). Does f(N)/q(N) tend to zero as N tends to infinity, where q(N) is the number of squarefree numbers <= N?