Why is "$1/\mathrm{Hz}$" unchanged while converting$\mathrm{V}^2/\mathrm{Hz}$ to $\mathrm{dB}/\mathrm{Hz}$?
Say you have a measurement in unit $U$ (V$^2$) per unit $u$ (Hz). I do claim (although not in public) that the logarithm is a kind of unitless transformation, see for instance What is the logarithm of a kilometer? Is it a dimensionless number? In other words, it is somehow $0$-homogeneous, while a power is $2$-homogeneous: multiplying by a constant yields a constant to some power.

First, the logarithm is  kind of homogeneous to a $0$-th power: it is the integral of $1/x$ (power $-1$). The derivative of an affine function is a constant. Derivatives decrease powers by one unit, integrals increase them of the same amount. The logarithm somehow stands with constants between $x^{-1}$ and $x^{1}$. Additionally, the coefficient of variation of $x^p$ when $p\to 0$ is about:

$$ \frac{x^p-x^0}{p-0}\approx \log x\,.$$

Second, a dB scale is often relative: to some signal measurement, to some "absolute" value. The $\log$ argument is better unitless: something in U divided by some unit $1$ U, called U$_0$. Thus, computing $\log(\mathrm{U}/\mathrm{U}_0)$ could be considered.

So  converting U/Hz into dB$_\text{U}$/Hz, to me, only affects the numerator.