Let $\delta(\phi) = \phi(0)$ be the dirac delta.
I would like to compute $\int_{\mathbb{R}} h(x) \delta(\lambda x) dx$
1) Since $\delta$ is an unit mass on $0$ $$\int_{\mathbb{R}} h(x) \delta(\lambda x) dx = h(0)$$
2)Before integrating against $\delta$ we change variables: $$\int_{\mathbb{R}} h(x) \delta(\lambda x) dx = \int_{\mathbb{R}} h(\frac{x}{\lambda}) \delta(x) \frac{dx}{\lambda} = \frac{h(0)}{\lambda}$$
Which way is correct? why? Is there an intuition why the correct answer is the correct answer? (for instance exploring the fact that $\delta = \lim_n f_n$ where $f_n$ is an approximation of unity)
Generally, what do we mean by $T(\lambda x)$ when $T$ is a distribution? To define $T$, we must specify what it does to test functions. This must be done in a way that is consistent with what happens when $T$ is an ordinary function. In the latter case $$ \int T(\lambda x)\phi (x)\,dx = \int T(y)\phi (\lambda^{-1} x)\,\lambda^{-1} dx = \lambda^{-1} \langle T, \phi(\lambda^{-1}y)\rangle $$ That is, the value $T(\lambda x)$ assigns to $\phi$ is equal to $\lambda^{-1}$ times the value that $T$ assigns to $\phi(\lambda^{-1} y)$. This is the desired definition of rescaled distribution $T(\lambda)$.
Apply the above to the case $T=\delta$, meaning that $\langle T,\phi\rangle =\phi(0)$. The result is that $\delta(\lambda x)$ takes the value $\lambda^{-1} \phi(0)$ on the test function $\phi$.