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No, even if you assume that $f$ is convex on $\mathbb R^{d+n}$. Take $n=2$ and let $y_1,\dots,y_{d+1}\in\mathbb R^2$ be vertices of a convex polygon. Let $X_1,\dots,X_{d+1}\subset\mathbb R^n$ be your …

Unless I misunderstood the question, here is a counter-example: $d=3$, $k=2$, $f:\mathbb R^3\to\mathbb R^2$ is just a linear map, e.g. $f(x,y,z)=(x,y)$. Take $q=(0,0)$, then $Y=f^{-1}(q)$ is a straigh …

EDIT: This answer is wrong: as pointed out by Alfonz, this map is not algebraic. The similar one with a rational approximation of $\sqrt2$ is, but then the degree is unbounded. Consider the following …

The answer is infinity for $n>2$. Suppose that there is an $i$ such that $T^i(x)=0$ for all integer vectors $x$. Then the same follows for all rational vectors by homogenuity, and then for all real v …

In general, no. Let $G$ be the group of all upper-triangular matrices with positive diagonal entries. It acts on $\mathbb R^2$ as a subgroup of $GL(2,\mathbb R)$. Consider $x=(1,0)$. Its orbit is the …