automorphisms or the group of scaling symmetries. Note that for $(\lambda_0,\ldots, \lambda_n)$ to be an element of ${\text{Aut}}(F_A)$, each monomial $\prod_{j=0}^n x_j^{a_{ij}}$ must be invariant under the action of $(\lambda_0, \ldots, \lambda_n)$.

Using the classification of Kreuzer and Skarke (see Remark 2.1), we can see that for any polynomial of one of the atomic types that each $\lambda_i$ must have modulus 1. If the polynomial $F_A$ is of fermat-type, then $\lambda^ax^a = x^a$ hence $\lambda^a = 1$. If $F_A$ is of loop-type, then $\lambda_i^{a_i}\lambda_{i+1} = 1$ for all $i < a_m$, hence $\lambda_{i+1} = \lambda_i^{-a_i}$. Moreover, $\lambda_m^{a_m} \lambda_1 = 1$ hence $\lambda_1 = \lambda_m^{-a_m} = \lambda_{m-1}^{a_ma_{m-1}} = \cdots = \lambda_1^{(-1)^ma_1\cdots a_m}$. If $|\lambda| \neq 1$ then $(-1)^m a_1\ldots a_m = 1$. This would require $m$ to be even and $a_i$ to be $1$ for all $i$. However, then the degree of the polynomial, $d$, must be $q_1 + q_2$; however $d = \sum_{i=0}^n q_i$, $n \geq 2$, and $q_i >0$, hence a contradiction is reached. Lastly, if $F_A$ is of chain-type, $\lambda_m^{a_m}x_m^{a_m} = x_m^{a_m}$, hence $|\lambda_m|^{a_m} = 1$. This implies that $|\lambda_{m-1}^{a_{m-1}}\lambda_m| = |\lambda_{m-1}^{a_{m-1}}| = 1$, and so on, hence $|\lambda_i| = 1$. Any polynomial that is a combination of 