I understand the linear transformation property of Gaussian distribution, i.e., if $x\sim\mathcal{N}(\mu, \sigma^2)$, then $cx\sim\mathcal{N}(c\mu, c^2\sigma^2)$ given $c$ is a constant.

If we write this as a marginalization:

$$p(f(c)|c) = \int p(f(c)|c, x)p(x)dx$$

and here $f(c) = cx$. $p(x)$ can be expressed as $\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2}(x-\mu)^2}$.

But how do we obtain $p(f(c)|c)$ from the above marginalization form (or the posterior predictive distribution, here I simplified the posterior to just $p(x)$), that is to get $$\quad \, p(f(c)|c) = \dfrac{1}{\sqrt{2\pi}c\sigma}e^{-\frac{1}{2c^2\sigma^2}(f(c)-\mu c)^2} \quad\text{?}$$

What troubles me is $p(f(c)|c, x)$.

I think it is $1$, as it is deterministic when $c$ and $x$ are given, but I'm not sure how to proceed.