Example 4 What positive integers $x$ satisfy $\varphi(2 x)=\varphi(3 x)$.
Let $x=2^{a} 3^{b} y$, where $a, b$ are non-negative integers, $6 \times y$.
If $b>0$, then
$$
\begin{array}{l}
\varphi(2 x)=\varphi\left(2^{a+1}\right) \cdot \varphi\left(3^{b}\right) \cdot \varphi(y)=2^{a} \cdot 3^{b-1} \cdot 2 \varphi(y), \\
\varphi(3 x)=\varphi\left(2^{a}\right) \cdot \varphi\left(3^{b+1}\right) \cdot \varphi(y)=2^{a-1} \cdot 3^{b} \cdot 2 \varphi(y) .
\end{array}
$$

Therefore, $\varphi(2 x) \neq \varphi(3 x)$. So, when $\varphi(2 x)=\varphi(3 x)$, $b=0, x=2^{a} y$.
At this time, $\varphi(2 x)=2^{a} \cdot \varphi(y), \varphi(3 x)=2 \varphi\left(2^{a}\right) \cdot \varphi(y)$.
Therefore, $\varphi\left(2^{a}\right)=2^{a-1}, a>0$.
Hence, $x=2^{a} y, a$ is a positive integer, $6 \nmid y$.