= Solution
For <multi-indices> $\alpha,\beta$, define
$$
p_{\alpha,\beta}(\varphi)=\sup_{x\in\mathbb R^n}|x^\alpha\partial^\beta\varphi(x)|,\qquad
\mathcal S(\mathbb R^n)=\{\varphi\in C^\infty:p_{\alpha,\beta}(\varphi)<\infty\ \text{for all }\alpha,\beta\}.
$$
These <seminorms> define the <Fréchet space> topology of the <Schwartz space>: $\varphi_j\to\varphi$ exactly when every $p_{\alpha,\beta}(\varphi_j-\varphi)$ tends to zero. An equivalent increasing family is
$$
q_m(\varphi)=\max_{|\beta|\leq m}\sup_x(1+|x|)^m|\partial^\beta\varphi(x)|.
$$
A <tempered distribution> is a continuous <linear functional> on this space. Equivalently, $|\langle T,\varphi\rangle|\leq Cq_m(\varphi)$ for some $C,m$. The usual <weak convergence of tempered distributions> means $\langle T_j,\varphi\rangle\to\langle T,\varphi\rangle$ for every fixed <Schwartz function>. The <strong dual topology> instead requires uniform convergence on every subset of the <Schwartz space> that is a <bounded set in a topological vector space>; the Fourier maps below are continuous in both topologies.
Using $\widehat\varphi(\xi)=\int e^{-ix\cdot\xi}\varphi(x)\,dx$, differentiation under the integral and <integration by parts> give
$$
\left|\xi^\alpha\partial_\xi^\beta\widehat\varphi(\xi)\right|
\leq\left\|\partial_x^\alpha(x^\beta\varphi)\right\|_{L^1}.
$$
The omitted coefficient has modulus one. By the <Leibniz rule>, every integrand is a finite sum of a polynomial times a derivative of $\varphi$. Inserting the integrable weight $(1+|x|)^{-n-1}$ bounds its <L1 norm> by finitely many <Schwartz space> seminorms. In particular $q_m(\widehat\varphi)\leq C_mq_{m+n+1}(\varphi)$. Thus the <Fourier transform> maps $\mathcal S$ continuously into itself.
For completeness, the <Fourier inversion theorem> follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi|^2}$ is $g_t(x)=(4\pi t)^{-n/2}e^{-|x|^2/(4t)}$. The <Fourier transform of a Gaussian> and <Fubini's theorem> show
$$
(2\pi)^{-n}\int e^{ix\cdot\xi}e^{-t|\xi|^2}\widehat\varphi(\xi)\,d\xi
=(g_t*\varphi)(x).
$$
As $t\downarrow0$, the right side tends to $\varphi(x)$ by the <approximate identity> property, while the left side tends to the undamped inverse integral by <dominated convergence theorem>, since $\widehat\varphi\in L^1$. Hence
$$
\boxed{\mathcal F^{-1}\psi(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}\psi(\xi)\,d\xi,\qquad
\mathcal F^2\varphi(x)=(2\pi)^n\varphi(-x).}
$$
The inverse is $(2\pi)^{-n}$ times reflection composed with the continuous <Fourier transform>, and is therefore continuous on the <Schwartz space>. This proves a continuous linear isomorphism with continuous inverse.
Define the <Fourier transform of a tempered distribution> by
$$
\langle\widehat T,\varphi\rangle=\langle T,\widehat\varphi\rangle.
$$
The continuous map on <Schwartz space> makes this a <tempered distribution>; the transpose of $\mathcal F^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the <strong dual topology>, the transform of a bounded set of <Schwartz functions> is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.
Now let $R\in SO(n)$ be a <rotation matrix> in the <special orthogonal group> and write $\rho_RT=T\circ R$. A change of variables with unit Jacobian gives
$$
\widehat{\varphi\circ R^t}(\xi)=\widehat\varphi(R^t\xi).
$$
The <rotation equivariance of the Fourier transform> on <tempered distributions> follows by duality:
$$
\begin{aligned}
\langle\widehat{\rho_RT},\varphi\rangle
&=\langle T,\widehat\varphi\circ R^t\rangle
=\langle T,\widehat{\varphi\circ R^t}\rangle\\
&=\langle\widehat T,\varphi\circ R^t\rangle
=\langle\rho_R\widehat T,\varphi\rangle.
\end{aligned}
$$
Therefore $\widehat{\rho_RT}=\rho_R\widehat T$. Applying this identity and the inverse <Fourier transform> gives the two directions:
$$
\boxed{T\circ R=T\ \text{for every }R\in SO(n)
\ \Longleftrightarrow\ \widehat T\circ R=\widehat T\ \text{for every }R\in SO(n).}
$$
This is precisely preservation of <radial tempered distributions>. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary <radial function>.
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