It is trivial to construct vertex-transitive polytopes: choose any finite matrix group $\Gamma\subseteq\mathrm O(\Bbb R^d)$ and some point $v\in\Bbb R^d$. Take the convex hull of the orbit $\Gamma v$, then this defined the orbit polytope
$$\mathrm{Orb}(\Gamma,v):=\mathrm{conv}(\Gamma v),$$
which is vertex-transitive (i.e. for any two vertices there is a symmetry of the polytope mapping the first vertex to the second).
Question: Is there a simiar way to generate an abundance of vertex-transitive tilings of $\Bbb R^d$?
By tiling I mean a face-to-face tiling of $\Bbb R^d$ by convex polytopes.
Update
I just had the following idea: take any lattice $\Lambda\subseteq\Bbb R^d$ and consider the Vornoi decomposition of $\Bbb R^d$ w.r.t this lattice. It is cell-transitive. So we compute its dual, which should then be vertex-transitive. Do we get the dual as the Vornoi decomposition of $\Bbb R^d$ w.r.t the vertices of the first Vornoi decomposition? I am not sure.
As you noted in the comments, the problem with $\mathbb R^d$ is that it's flat, and the convex hull of a set of points "forgets about" the interior points.
Consider the mapping from $\mathbb R^d$ to a paraboloid in $\mathbb R^{d+1}$ defined by
$$x=x_1e_1+\cdots+x_de_d\mapsto$$
$$x^\smile=x+\tfrac12x^2e_{d+1}$$
$$=x_1e_1+\cdots+x_de_d+\tfrac12(x_1^2+\cdots+x_d^2)e_{d+1}$$
where $e_i$ are basis vectors. This gives us a curved model of Euclidean space. It's not intrinsic curvature, though; if we use a degenerate dot product on $\mathbb R^{d+1}$, with $e_i\cdot e_j=(1-\delta_{i,d+1})\delta_{i,j}$, then the resulting geometry on the paraboloid is ordinary Euclidean geometry.
Furthermore, translations, rotations, reflections, etc. can be induced by affine transformations on $\mathbb R^{d+1}$. We need only consider reflections, since other Euclidean transformations are compositions of these. Let $y\in\mathbb R^d$ be any point, and $u\in\mathbb R^d$ any unit vector, normal to a hyperplane passing through $y$. Then the reflection of $x\in\mathbb R^d$ across this hyperplane is
$$x\mapsto$$
$$x-2\big((x-y)\cdot u\big)u.$$
Now consider this affine transformation on $\mathbb R^{d+1}$:
$$x+x_{d+1}e_{d+1}\mapsto$$
$$x+x_{d+1}e_{d+1}-2\big((x-y)\cdot u\big)\big(u+(y\cdot u)e_{d+1}\big).$$
(Notice that $u+(y\cdot u)e_{d+1}=u_y^\smile$ is a tangent vector to the paraboloid at the point $y^\smile$, and that factor in the middle is $(x-y)\cdot u=(x+x_{d+1}e_{d+1}-y^\smile)\cdot u_y^\smile$ using the degenerate dot product.)
Its restriction to the paraboloid is
$$x^\smile\mapsto$$
$$x+\tfrac12x^2e_{d+1}-2\big((x-y)\cdot u\big)\big(u+(y\cdot u)e_{d+1}\big)$$
$$=\Big(x-2\big((x-y)\cdot u\big)u\Big)+\tfrac12\Big(x-2\big((x-y)\cdot u\big)u\Big)^2e_{d+1}$$
$$=\Big(x-2\big((x-y)\cdot u\big)u\Big)^\smile$$
which has the same effect as the reflection on $\mathbb R^d$.
Using this, you should be able to express any Euclidean isometry group in terms of these non-Euclidean reflections on $\mathbb R^{d+1}$. Then you can take the convex hull of the orbit of any point $x^\smile$, find the hull's faces, and project these back onto $\mathbb R^d$.
Note that the convex hull of a set $S$ has $F(\text{conv}(S))=\text{conv}(F(S))$ for any affine transformation $F$, so it doesn't matter where the origin is in $\mathbb R^d$, or where the vertex of the paraboloid is.
Since the paraboloid is strictly convex, no points are "forgotten" (absorbed into the convex hulls of other points).
Any face of the convex hull which is unbounded must be vertical (its covector must have no $e^{d+1}$ component). Equivalently, any non-vertical face is bounded. For, if a face is defined by $\alpha_1x_1+\cdots+\alpha_dx_d+x_{d+1}=b$, since the convex hull is above the paraboloid, we have $x_{d+1}\geq\tfrac12x^2$ and thus $\alpha_1x_1+\cdots+\alpha_dx_d+\tfrac12x^2\leq b$, which describes a ball in $\mathbb R^d$. So $x_1,\cdots,x_d$ are bounded, and it follows that $x_{d+1}=b-\alpha_1x_1-\cdots-\alpha_dx_d$ is also bounded.
With a vertical face, all of the points in $\mathbb R^d$ are on one side of a hyperplane. So if your chosen group is cocompact, then any point's orbit crosses any hyperplane, and each face must be bounded.
If your chosen group is discrete, then each bounded face must be the convex hull of a finite set of points. (Otherwise the points would accumulate and not be discrete.)