Let $\phi \colon \mathbb{R}^n \to \mathbb{R}$ be continuous with compact support. Furthermore, suppose that $\phi$ is spherically symmetric (i.e., assume $\phi(Tx) = \phi(x)$ for every orthogonal transformation $T\colon\mathbb{R}^n \to \mathbb{R}^n$).
Next, for $(x,t) \in \mathbb{R}^n \times (0, \infty)$, let $G(x,t) = (4\pi t)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4t}}$ be the fundamental solution to the heat equation. Prove that the convolution $f(x,t) = \displaystyle \int_{\mathbb{R}^n} G(y, t)\phi(x-y)dy$ is also spherically symmetric in the variable $x \in \mathbb{R}^n$. In other words, we want to show that, for any $t > 0$, and any orthogonal $T\colon\mathbb{R}^n \to \mathbb{R}^n$, we have $f(Tx, t) = f(x, t)$.
This question appears on an old PDE qual I am studying, and I'm stumped so far. One thing I have noticed is that the function $G$ is spherically symmetric in the variable $x$. So, I'm wondering if there is a more general result I could apply, something like: "the convolution of two integrable, spherically symmetric functions on $\mathbb{R}^n$ is again spherically symmetric."
I also know that the convolution is a commutative operation, so I could just as well define $f$ by $f(x,t) = \displaystyle \int_{\mathbb{R}^n} G(x-y, t)\phi(y)dy$. But I am failing to see how that would make the proof easier.
Hints or solutions are greatly appreciated.