I have a standard Gaussian random variable $Z$ in $\mathbb{R}^d$ and a nonsingular linear map $T(x)=Mx$ which fixes a $(d-1)$ subspace, say $\{x_1=0\}$. That is,
$$M=\left(\begin{array}{cc}M_{11} & 0\\ M_{\cdot 1} & I_{(d-1)\times(d-1)}\end{array}\right)$$
Let $K$ be a convex set (nonempty interior) contained in $\{x_1\ge 0\}$. If it helps, K is also a cone generated by $d$ linearly independent vectors.
I use $\mathbb{1}_K$ for the indicator function.
Question: Suppose $\mathbb{E}(Z\mathbb{1}_{T(K)})=c\mathbb{E}(Z\mathbb{1}_K)$ for some constant $c$. Must $M$ be a diagonal matrix? Moreover, if $K$ is a cone generated by $d$ linearly independent vectors, must $T$ be the identity map?
Progress: Let $\psi$ be the standard gaussian measure, $\psi^\ast$ the pull-back through $T$, and perform a change of variables to get
$$\mathbb{E}(Z\mathbb{1}_{T(K)})=\int_{T(K)}x d\psi(x)=\int_K T(x) d\psi^\ast(x)$$
Where $\psi^\ast$ has density proportional to $$e^{-|T(x)|^2/2}$$
I don't know if this is useful or where to go from here. I would give a big hug for any help. Thanks!
As you have stated it, $T$ does not need to be the identity.
Consider any $K$ where the cross-sections perpendicular to $x_1$ are centered on the $x_1$ axis. Maybe you can use the idea of a symmetral where you slide the cross sections parallel to each other to achieve this centeredness. Then, $M$ can be any diagonal matrix in addition to your restrictions and satisfy your requirements.