How to prove $\delta(kx) = \frac{\delta(x)}{|k|}$ by using properties of a test function. So, I have a Fourier analisys course right now and got a problem to solve. Here's how it sounds

Given that $\delta(x)$ is a Dirac delta function, $\phi(x)$ is a test function, by using properties of a test function prove that $\delta(kx) = \frac{\delta(x)}{|k|}$. (Hint: $\langle \delta(x), \phi(x) \rangle = \phi(0)$)

And, here's how I tried to prove that. I assumed that hint is not for nothing there, so I decided that it should be a decent place to start.
$$ \langle \delta(x), \phi(x) \rangle = \phi(0) $$
$$ \langle \delta(kx), \phi(kx) \rangle = \phi(0) $$
$$ \int_{-\infty}^{\infty} \delta(kx) \phi(kx) d(kx) = \phi(0) $$
$$ u = kx \implies du = d(kx) = kdx $$
$$ \int_{-\infty}^{\infty} k \delta(kx) \phi(kx) dx = \phi(0) $$
$$ \int_{-\infty}^{\infty} \delta(kx) \phi(kx) dx = \frac{\phi(0)}{k} $$
And... I am not sure what exactly I achieved here. $ \delta(kx) $ is still $ \delta(kx) $, so I assume I made no progress by going this way, and starting with a hint was not a good idea. So I started over with a slightly different approach.
$$  \langle \delta(kx), \phi(x) \rangle = $$
$$  = \int_{-\infty}^{\infty} \delta(kx) \phi(x) dx = * $$
$$  u = kx \implies  du = kdx $$
$$  * = \int_{-\infty}^{\infty} \frac{1}{k} \delta(u) \phi(\frac{u}{k}) du = ??? $$
A-a-and, I'm in another dead end again. I feel that I need to somehow utilize the hint $\langle \delta(x), \phi(x) \rangle = \phi(0)$ and it should somehow lead me to $ \int_{-\infty}^{\infty} \frac{1}{k} \delta(u) \phi(\frac{u}{k}) du = \frac{\phi(0)}{k} $ which I'm not even sure is true or not, but even if it is, it still does not put me any closer to the statement I need to prove and absolutely does not use any properties of a test function.
I understand that this question is probably a duplicate, and I already checked a lot of answers like this one, but, sadly, they don't really help with my problem because they don't use the definition and properties of a test function, but the definition of a Dirac delta itself, which I believe is not what this question about. Or they actually do, but I can not understand in what way.
So, please, I clearly do not understand something about the question and would be very thankful if someone helped me solve this problem.
 A: Yes, this is a chronically possibly-counterintuitive aspect of $\delta$. But there is a two-fold way to understand it.
First, somewhat the intention of $\delta$, is as some kind of function on $\mathbb R$ (for example) that "can be integrated against" very nice functions $\varphi$, and $\int \delta(x)f(x)dx=f(0)$. Whether or not this is literally correct/rigorous, it certainly suggests the heuristic that, for $k>0$,
$$
\int \delta(kx)\,f(x)\;dx \;=\; \int \delta(x)\,f(x/k)\;d(x/k)
\;=\; {1\over k} \int \delta(x)\,f(x/k)\;dx
\;=\; {1\over k} f(0/k) \;=\; {1\over k}f(0)
$$
(Yes, there is a little fooling around to see that it's $|k|$ for $k<0$, and not just $k$, but this is a subsidiary point... $\delta$ is even...)
A disquieting aspect of that argument is that $\delta$ is not literally a function, and that integral cannot be literal. But the heuristic is what we want to be correct. The change of variables is certainly correct for any even function in place of $\delta$... and we do want $\delta$ to be some kind of limit of classical, nice functions, so we insist that the conclusion is correct (for consistency with classical functions, and expressing $\delta$ as a weak limit of them!).
So, second, on $\mathbb R^n$ (just to see where the dimension enters), we define the dilation $u(kx)$ with $k>0$ of a distribution $u$ (even though it's not a pointwise-valued function... so writing this as though $u$ had an argument is misleading) for $k>0$ by $u(kx)(f)=k^{-n}u(f(x/k))$. More precisely, letting $T_ku$ be that dilation, without mentioning pointwise values, $(T_ku)(f)=k^{-n}\cdot u(T_{k^{-1}}f)$.
