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$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) $$ for real $x$, where $\text{erfi}(x):=\sqrt{\frac2\pi}\int_0^{x\sqrt2}e^{z^2/2}\,dz$.

Here are the graphs of $u_1$ (blue) and $\Re u_2$ (gold) over the interval $[-6,6]$:

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) \\ =\frac54\,u_1(x)-\frac{3ix}{4\sqrt{2\pi}}\,e^{-x^2/2} $$ for real $x$, where $\text{erfi}(x):=\sqrt{\frac2\pi}\int_0^{x\sqrt2}e^{z^2/2}\,dz$.

Here are the graphs of $u_1$ (blue) and $\Re u_2$ (gold) over the interval $[-6,6]$:

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) $$ for real $x$.

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) $$ for real $x$, where $\text{erfi}(x):=\sqrt{\frac2\pi}\int_0^{x\sqrt2}e^{z^2/2}\,dz$.

Here are the graphs of $u_1$ (blue) and $\Re u_2$ (gold) over the interval $[-6,6]$:

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{(2-x^2)}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) $$ for real $x$.

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.

Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.

Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.

E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=|t|e^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=e^{-t^2/2}\,t\,(2\,1(t>0)-\tfrac12\,1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.

The corresponding $A$ and solutions $u_1,u_2$ of \eqref{1} are given by $$A(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat A(t)\,dt= \frac{2-x^2}{8 \sqrt{\pi }}\,e^{-x^2/4}, $$ $$u_1(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_1(t)\,dt =\frac{1}{\pi } -\frac 1{\sqrt{2 \pi }} e^{-x^2/2} x\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right),$$ $$u_2(x)=\frac1{2\pi}\,\int_\R e^{-itx} \hat u_2(t)\,dt =\frac{5}{4 \pi }-\frac x{4 \sqrt{2 \pi }}e^{-x^2/2} \left(3 i+5\, \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right) $$ for real $x$.