Given $F$ is any square matrix, let $F^T$ be the conjugate transpose.
Is it possible that $\operatorname{rank}(F F^T-F^T F) = 1?$?
It seems that it is impossible, but I don't know how to prove it.
On the other hand, is there specific name for the matrix in the form of $F F^T-F^T F$?
since you mention conjugate transpose, I infer that the field is $\mathbb C$. The answer then is no.
$A:= FF^* - F^*F$
$A$ is Hermitian and hence diagonalizable. This means $\text{rank}\big(A\big)=1$ can only occur because $\lambda_1 \neq0$ and $\lambda_k = 0$ for $k\in\big\{2,...,n\big\}$, but supposing this is true
$\lambda_1 $
$= \lambda_1 + \big(\sum_{k=2}^n \lambda_k\big)$
$= \text{trace}\big(A\big) $
$= \text{trace}\big(FF^*-F^*F\big)$
$= \text{trace}\big(FF^*\big)-\text{trace}\big(F^*F\big)$
$=\text{trace}\big(FF^*\big)-\text{trace}\big(FF^*\big)$
$=0$
which is a contradiction