How can I show if $AA^T+A^TA=A+A^T$, then $Col(A)=Col(A^T)$?

We prove that the null-spaces of $A$ and $A^T$ are equal by showing $$ Av=0\implies A^Tv=0 \\A^Tv=0\implies Av=0 $$ First, to show $Av=0\implies A^Tv=0$, we assume $Av=0$ for some vector $v$. Then by multiplying both sides of $AA^T+A^TA=A+A^T$ in $v^T$ from left and $v$ from right we obtain $${ v^TAA^Tv+v^TA^TAv=v^TAv+v^TA^Tv\implies \\ v^TAA^Tv+v^TA^TAv=2v^TAv\implies \\ v^TAA^Tv+0=0\implies \\ ||A^Tv||^2=0\implies \\ A^Tv=0 }$$ Similarly the proof holds conversely and the statement is proved $\blacksquare$.

Using your observation that the null space is the orthogonal complement to the row space, it suffices to show that $\operatorname{Null}(A) = \operatorname{Null}(A^T)$. Now, decompose $A$ into the sum $A = S + K$, where $S$ is symmetric and $K$ is skew-symmetric. In fact, this implies that $S = \frac 12(A + A^T)$ and $K = \frac 12 (A - A^T)$.

Suppose that $x \in \Bbb R^n$ is such that $Ax = 0$. We have $$ Ax = 0 \implies (S + K)x = 0 \implies Sx = -Kx. $$ It follows that $x^TSx = -x^TKx$. Because $K$ is skew-symmetric, we must have $x^TKx = 0$ (why?). Thus, we have $x^TSx = 0$. That is, $$ \begin{align} 0 = x^TSx &= \frac 12 x^T[AA^T + A^TA]x = \frac 12 [x^TAA^Tx + x^TA^TAx] \\ & = \frac 12 [(A^Tx)^T(A^Tx) + (Ax)^T(Ax)] = \frac 12 [\|A^Tx\|^2 + \|Ax\|^2]. \end{align} $$ Thus, we must have $\|A^Tx\| = 0$, so $A^Tx = 0$.

That is, we have $Ax = 0 \implies A^Tx = 0$, so that $\operatorname{Null}(A) \subseteq \operatorname{Null}(A^T)$. Symmetrically, we can conclude that $\operatorname{Null}(A^T) \subseteq\operatorname{Null}(A)$. So, we have $\operatorname{Null}(A) = \operatorname{Null}(A^T)$ as desired.