Let $Z=\{U\in M_n(\mathbb{R}); U \text{ is symmetric and }\operatorname{tr}(U)=0\}$. $Z$ is an algebraic set of dimension $\dfrac{n^2 +n-2}{2}$.
We consider the algebraic function $f:X\in M_n(\mathbb{R})\mapsto XX^T-X^TX\in Z$.
$\textbf{Question 1}$. Is $f$ surjective ?
The answer seems to be yes. We consider the equation
$(*)$ $XX^T-X^TX=A$ where $A\in Z$.
Clearly, we may assume that $A$ is diagonal with $\sum_i a_{i,i}=0$; we may also assume that, for every $i$, $a_{i,i}\not= 0$ (otherwise, we can reduce the problem to a matrix of dimension $<n$).
I did not find any counterexample (using random matrices $A$ and numerical calculations up to $n = 10$).
$\textbf{Question 2.}$ What is the dimension of the fiber $f^{-1}(A)$, that is to say, what is the number of parameters on which the general solution (in $X$) of the equation $(*)$ depends ?
- If $A=0$, then $X$ is normal and the required dimension is $\dfrac{n(n+1)}{2}$.
EDIT.
- Now we assume that $A$ is INVERTIBLE. It seems that $f$ is a submersion in a solution $X$ and that the dimension is $n^2-\dim(Z)=\dfrac{n^2-n+2}{2}$. At least, that's true for the examples I tested ($n\leq 4)$.
$\textbf{Remark}$. The above conjecture does not work when $\mathrm{rank}(A)<n$. Two examples
i) when $A=\mathrm{diag}(1,1,-2,0)$ a solution of $(*)$ is $X_0=\begin{pmatrix}0&\pm 1&0&0\\0&0&\pm\sqrt{2}&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$ (cf. user1551's answer); then $\dim(\ker(Df_{X_0}))=8$ - instead of $\dfrac{n^2-n+2}{2}=7$ - and $\dim(f^{-1}(A))\geq 8$.
ii) when $A=\mathrm{diag}(1,-1,0,0)$ a solution is $X_1=\begin{pmatrix}0&\pm 1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$; then $\dim(\ker(Df_{X_1}))=10$ and $\dim(f^{-1}(A))\geq 10$.
For your first question, note that for every vector $z$ whose entries sum to zero, we may pick a sufficiently large number $C>0$ and write $z=(I-P)y$ for some permutation matrix $P$ and for positive vector $y$: $$ \pmatrix{z_1\\ z_2\\ \vdots\\ z_{n-1}\\ -\sum_{i=1}^{n-1}z_i} =\pmatrix{1&&&&-1\\ -1&\ddots\\ &\ddots&\ddots\\ &&\ddots&\ddots\\ &&&-1&1}\pmatrix{z_1+C\\ z_1+z_2+C\\ \vdots\\ z_1+\cdots+z_{n-1}+C\\ C}. $$ Let $x$ be the entrywise square root of $y$. For any real orthogonal matrix $Q$, let $X=Q\operatorname{diag}(x)P^TQ^T$. Then $$ XX^T-X^TX =Q\left(\operatorname{diag}(y)-P\operatorname{diag}(y)P^T\right)Q^T =Q\operatorname{diag}(z)Q^T. $$ Hence $f$ is surjective.