For every simplex $\Delta\subset\Bbb R^n$ with $n+1$ vertices and every number $m\le n$ there is an $m$-dimensional subspace $W\subseteq\Bbb R^n$ so that the orthogonal projection of $\Delta$ onto $W$ has $n+1$ vertices.
Proof.
Fix a polytope $P\subset\Bbb R^m$ with $n+1$ vertices.
First, we show that $P$ is the orthogonal projection of some simplex $\Delta_P \subset\Bbb R^n$: if $P$ has vertices $v_0,...,v_n\in\Bbb R^m$, then for $i\in\{0,...,n\}$ the
$$w_i:=\underset{\!\!\!\!\!\!\!\!\begin{array} .\uparrow\\ \!\!\!\!\!\!\!\!\text{at index $m$}\end{array}}{\begin{cases}
(v_i,0,...,0) & \text{for $i\le m$} \\
(v_i,0,...,1,...,0) & \text{for $i > m$}
\end{cases}}\quad\in\Bbb R^n$$
are $n+1$ points in $\Bbb R^n$ in convex position.
They necessarily form the vertices of some simplex $\Delta_P$.
Also, the projection onto the first $m$ coordinates (which is orthogonal) yields the initial polytope $P$.
Alternatively, we can say that the intersection of the subspace $\Bbb R^m\subset\Bbb R^n$ with the cylinder $C:=P+\Bbb R e_{m+1}+...+\Bbb R e_n$ is $P$.
Now, any two simplices in $\Bbb R^n$ with $n+1$ vertices are affinely equivalent, and so we can choose an affine transformation $X$ with $X\Delta_P= \Delta$. Clearly, the intersection of the transformed cylinder $X C$ with the transformed subspace $W':=X\Bbb R^m$ is still a polytope with $n+1$ vertices. The same is true if you replace $W'$ by a subspace $W\subseteq\Bbb R^n$ orthogonal to $Xe_i,i\in\{m+1,...,n\}$, as all cross-sections of the cylinder are affinely equivalent. But this intersection can be interpreted as the orthogonal projection of $\Delta=X\Delta_P$ onto $W$, which then has $n+1$ vertices. $\quad$
$\square$
As a side note, the intersection $XC\cap W$ is an affine transformation of $P$. Thus, given any simplex $\Delta\subset\Bbb R^n$ and any polytope $P\subset\Bbb R^m$ with $n+1$ vertices, $P$ is affinely equivalent to an orthogonal projection of $\Delta$.