**Theorem**: The set of positive integers $n$ such that there is a $1$-step vanishing polyplet with $n$ cells is $$ \{1,2,9,10,12,14 \} \cup  \mathbb{N}_{\ge 15}$$ It can be represented by $1$-step vanishing [polyominoes][1] without hole and a non-trivial symmetry group.   

*Proof*: The finite part is given by:  
[![enter image description here][2]][2]  
For the infinite part, we will do three steps.
Firstly, the family beginning as follows:  
    
[![enter image description here][3]][3]   

provides the integers $6a+b$ with $b \in \{ 3 , \dots , a+2 \}$. Now, $$6a + (a+2) +1 \le 6(a+1)+3$$ if and only if $a \ge 6$. But $6 \times 6+3  = 39$, so it is ok for $\mathbb{N}_{\ge 39}$.    
 
Next, the gaps of the previous family for $14<n<39$ are given by the set $$\{17,18,19,20,24,25,26,31,32,38\}$$ and can be filled ($n=19,20$ excepted) by the family beginning as follows:     

[![enter image description here][4]][4]   

providing the integers $7a+b$ with $b \in \{ 3 , \dots , a+2 \}$.    

Finally, $n=19,20$ are given by:   
 
[![enter image description here][5]][5]

The result follows. $\square$  

*Bonus question*: Can we classify (in some sense) the $1$-step vanishing [polyominoes][1] without hole?  

There are $21$ such polyominoes with $n$ cells and $n \le 18$.


  [1]: https://en.wikipedia.org/wiki/Polyomino
  [2]: https://i.sstatic.net/yAPao.png
  [3]: https://i.sstatic.net/BEegP.png
  [4]: https://i.sstatic.net/c39of.png
  [5]: https://i.sstatic.net/MUzqz.png