Is there an $n\geqslant 3$ and a symmetric, convex polytope $V$ in $\mathbb R^n$ (symmetric means that if $x\in V$ then $-x\in V$ too) such that
- $V$ has $2n$-vertices and
- $V$ is not image of the cross-polytope $$ \{x\in\mathbb R^n\colon |x_1| +\ldots + |x_n| \leqslant 1\}?$$ by a linear map?
No.
Let the vertices be $v_1,v_2,\cdots,v_n,\;v_{n+1}=-v_1,\;v_{n+2}=-v_2,\;\cdots,\;v_{2n}=-v_n$.
Suppose the vectors $v_1,v_2,\cdots,v_n$ are linearly dependent. Then one of them is in the span of the other $n-1$ vectors. It follows that $v_{n+1},v_{n+2},\cdots,v_{2n}$ are also in the span of those $n-1$ vectors, so the whole polytope is in an $(n-1)$-dimensional subspace. But you mentioned that the polytope must be full-dimensional. Hence, the vectors $v_1,v_2,\cdots,v_n$ must be linearly independent, and form a basis for $\mathbb R^n$.
Let $M$ be the linear map defined by
$$(1,0,0,\cdots,0,0)\mapsto v_1$$ $$(0,1,0,\cdots,0,0)\mapsto v_2$$ $$\vdots$$ $$(0,0,0,\cdots,0,1)\mapsto v_n$$
or equivalently the matrix
$$M=\begin{bmatrix}&&&\\v_1&v_2&\cdots&v_n\\&&&\end{bmatrix}.$$
Then $M$ maps the standard cross-polytope to your polytope.