If $\phi \in C_c^\infty(\mathbb R^n) \cap L^1(\mathbb R^n)$ such that $\| \phi \|_1 = 1$, then the dillations $\phi_t$ preserve these properties?

Context. Consider the usual Lebesgue measure on $\mathbb R^n$ and denote by $L^1(\mathbb R^n)$ the usual space of measurable functions that are integrable. Moreover, let $C^k(\mathbb R^n)$ denote the space of functions defined on $\mathbb R^n$ that are $k$ times continuously differentiable, where $k \geqslant 1$ is an integer. As usual, define $C^\infty(\mathbb R^n)$ to be the intersection of all $C^k$ and let $C_c(\mathbb R^n)$ denote the space of continuous functions on $\mathbb R^n$ with compact support. Furthermore, let $C_c^\infty(\mathbb R^n) = C_c(\mathbb R^n) \cap C^\infty(\mathbb R^n)$.

Consider an arbitrary kernel $\phi \in C_c^\infty(\mathbb R^n) \cap L^1(\mathbb R^n)$ such that $\| \phi \|_1 = 1$ and define the dillations

$$ \phi_t(x) = t^{-n} \, \phi\left( \frac{x}{t} \right), $$

for every $x \in \mathbb R^n$ and $t > 0$.

Question. My goal is to prove the following claim:

$$ \phi_t \in C_c^\infty(\mathbb R^n) \cap L^1(\mathbb R^n) \, \, \, \text{ such that } \, \, \, \| \phi_t \|_1 = 1, \quad \text{ for every }t > 0. $$

My attempt. I believe I was able to prove that $\phi_t \in L^1(\mathbb R^n)$ with $\| \phi_t \|_1 = 1$, for every $t > 0$. To do this, I applied a simple change of variables to obtain

$$ \| \phi_t \|_1 = \int_{\mathbb R^n} |\phi_t(x)| \, dx = \int_{\mathbb R^n} t^{-n} \Bigg| \phi\left( \frac{x}{t}\right) \Bigg| \, dx = \int_{\mathbb R^n} t^{-n}|\phi(u)|t^n \, du = \| \phi \|_1 = 1. $$

On the other hand, I am not sure how to proceed to establish that $\phi \in C_c^\infty(\mathbb R^n)$.

Thanks for any help in advance.