For what numbers $x$ is $D(x) = 2x - \sigma_1(x)$ equal to $\varphi(x)$?

Question

Let $\sigma_1$ be the classical sum-of-divisors function.

Let $$D(x) = 2x - \sigma_1(x)$$ be the deficiency of $x$.

Let $\varphi(x)$ denote the Euler totient.

Here is my question:

For what numbers $x$ is $D(x) = 2x - \sigma_1(x)$ equal to $\varphi(x)$?

Trivially, $x=1$ answers my question. But are there any others?