Is $Y$, obtained from a random uniform unitary, uniformly distributed?

Fix a vector $x \in \mathbb{C}^m$. Let $\mu$ be the surface measure on $U(m)$, and let $dS$ be the surface measure on $S = \{z \in \mathbb{C}^m : \lVert z \rVert = \lVert x \rVert\}$, both normalized to be probability measures. Let $\nu$ be the distribution of your random variable $y$, i.e. for Borel $E \subset S$, $\nu(E) = \mu(f^{-1}(E))$, where $f(g) = gx$. We will show that $\nu = dS$.

Let $u : S \to [0, \infty]$ be an arbitrary measurable function. We have \begin{align} \int_{S}u(s)\,d\nu(s) &= \int_{U(m)}u(gx)\,dg \\ \end{align}

Now I claim that $Au(x) = \int_{U(m)}u(gx)\,dg$ is a radial function, meaning that if $y = Tx$, with $T \in U(m)$, then $Au(x) = Au(y)$. With $y = Tx$, observe that $$Au(y) = \int_{U(m)}u(gTx)\,dg.$$ Note that the map $R_T : U(m) \to U(m)$ defined by $R_Tg = gT$ is linear and is an isometry (using the operator norm) since $\lVert gT \rVert = \lVert g \rVert$. Now let $\Omega \subset U(m)$ be open and parameterized by a chart $\phi : O \to \Omega$, where $O$ is open in $\mathbb{R}^{\dim U(m)}$. Suppose $f : U(m) \to [0, \infty]$ is a measurable function supported on $\Omega$. Then $f \circ R_T$ is supported on $R_T^{-1}\Omega$ and $\psi(z) = R_T^{-1}\phi(z)$ defines a chart $\psi : O \to R_T^{-1}\Omega$. Note that $D\psi(z) = R_T^{-1}D\phi(z)$, so $D\psi(z)^tD\psi(z) = D\phi(z)^tD\phi(z)$ since $(R_T^{-1})^t(R_T^{-1}) = I$. Thus \begin{align} \int_{U(m)}f(R_Tg)\,dg &= \int_{R_T^{-1}\Omega}f(R_Tg)\,dg \\ &= \int_{O}f(R_T\psi(z))\sqrt{\det D\psi(z)^tD\psi(z) }\,dz \\ &= \int_{O}f(\phi(z))\sqrt{\det D\phi(z)^tD\phi(z)}\,dz \\ &= \int_{\Omega}f(g)\,dg \\ &= \int_{U(m)}f(g)\,dg. \end{align} Since the above holds when $f$ is supported in a single coordinate patch, it holds for all $f$ by linearity since we can cover $U(m)$ by finitely many coordinate patches since it is compact. This proves in particular that $Au(y) = Au(x)$, so $Au$ is a radial function. Thus \begin{align} \int_{S}u(s)\,d\nu(s) &= \int_{U(m)}u(gx)\,dg \\ &= \int_{S}\int_{U(m)}u(gy)\,dg\,dS(y) \\ &= \int_{U(m)}\int_{S}u(gy)\,dS(y)\,dg \end{align} Now a similar argument to before shows that $\int_{S}u(gy)\,dS(y) = \int_{S}u(y)\,dS(y)$ since $g$ is a linear isometry. Hence $$\int_{S}u(s)\,d\nu(s) = \int_{U(m)}\int_{S}u(y)\,dS(y)\,dg = \int_{S}u(y)\,dS(y).$$ Since $u : S \to [0, \infty]$ was arbitrary, this means $\nu = dS$.