Compute $\phi_X(t)=E(e^{it^\top X})$ if $X\stackrel{d}{=}\mu + \xi AU$ with $AA^\top=\Sigma$

Sorry for asking, I’m just quite unsure;

So assume $X\stackrel{d}{=}\mu + \xi AU$, where $\mu$ is an $n$-dimensional constant vector, $AA^\top=\Sigma$, a constant $n \times n$ matrix, $\xi$ a 1-dimensional random variable and $U$ an $n$-dimensional random variable and $U$ and $\xi$ independent from each other.

Also of note is that for the characteristic function of U, it holds: $\phi_U(t)=\psi(t^\top t)$ for some function $\psi$.

Now I want to show that the characteristic function of $X$ has the form: $\phi_X(t)=e^{it^\top \mu}\psi(t^\top \Sigma t)$;

I start like that:

For the characteristic function of $X$ we have:
\phi_X(t) &=& \mathbb{E}\left( \exp(i t^\top(\mu + \xi AU) \right) =\exp(i t^\top \mu) \mathbb{E}\left(\exp(i t^\top\xi AU)\right)=\exp(i t^\top \mu)\mathbb{E}\left( \mathbb{E}\left( \exp(i t^\top\xi AU) | \xi \right) \right) \\
&=& \exp(i t^\top \mu) \mathbb{E}\left( \phi_{AU}\left( \xi t \right) \right),
where we used the law of total expectation. Now we have:
\phi_{AU}(t) = \mathbb{E}\left( \exp(i t^\top AU) \right)=\phi_U(A^\top t)=\psi_U((A^\top t)^\top A^\top t)=\psi_U(t^\top A A^\top t)= \psi_U(t^\top \Sigma t).

Is this correct, so far? What do I do now? The only thing I know about $\xi$ is that $\mathbb{E} \xi^2=n$ and the independence from $U$.

Anyone able to help me?

Oh of note is also, that the $\psi$ in the characteristic function of $U$ and the $\psi$ in the equation I want to prove don’t need to be the same

EDIT: If you don’t know how to continue, I’m also happy about anyone telling me if my calculation is correct or if there are some mistakes

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