Edit: Whuber answered it better: $AA^T$ and $A^TA$ have different dimensions, so they won't be equal
Your question basically asks is $A A^T = A^T A$? In general, the answer is no.
Define $C=AB$ for matrices $A,B$. The element at row $i$, column $j$ of $C$ is $C_{i,j}=\sum_k A_{i,k}B_{k,j}$, by the definition of matrix multiplication. If $AA^T=A^TA$, then element-wise they should be the same. Of course, A must have the same number of rows and columns, otherwise the two products have different dimensions, so the answer is no in that case. If $A$ is a square matrix, though, we can write the condition needed for the equality to hold:
If we let $C=AA^T$, then $C_{i,j}=\sum_k A_{i,k} (A^T)_{k,j}=\sum_k A_{i,k} A_{j,k}$ by the property of matrix transpose $A_{i,j} = (A^T)_{j,i}$.
Now, consider $C'=A^TA$. We have that $C'_{i,j}=\sum_k (A^T)_{i,k} A_{k,j}=\sum_k A_{k,i} A_{k,j}$.
Since, in general, $\sum_k A_{k,i} A_{k,j} \neq \sum_k A_{i,k} A_{j,k}$, we have that $A A^T \neq A^T A$ for $A$ a square matrix.
Some notable exceptions for which $A^TA=AA^T$ are the identity matrix and symmetric matrices where $A=A^T$, but since in PCA $A$ is usually a design matrix (rows are observations, columns are features), you won't be any of these exceptions.