Let $\varphi(x)$ be the Euler totient function. For $k = 1,2,3,\ldots$ I calculated the number of solutions of $\varphi(x) = \varphi(x+k)$. I observed that we have very few solutions when $k = 3,9,15,21,\ldots$ and a very large number of solutions when $k = 6,12,18,24,\ldots$ as shown below for $n \le 10^7$.
Question 1: Why do we have a very few solutions when $k$ is an odd multiple of $3$ and a very large number of solutions when $k$ is an even multiple of $3$?
Question 2: Is the following claim true?
For any six consecutive positive integer values of $k$, and a sufficiently large $x$, the number of integers $n \le x$ such that $\varphi(n) = \varphi(n+k)$, is the lowest when $k \equiv 3 \text{ (mod 6)}$ and is the highest when $k \equiv 0 \text{ (mod 6)}$.
Related question: Gap between integers having the same number of co-primes