I have a complex-symmetric matrix (in the sense $A=A^{T}$ not $A=A^{H}$), which is required to be positive semi-definite in the following sense (sometimes referred to as positive real):
$ \Re(x^{*} A x) \ge 0 $ for any $x$
This matrix is a projection matrix which is formed from the outer product of a vector with itself
$ A = v \otimes v$ where $v$ is a complex vector
I know that this is equivalent to requiring that $\Re(A)$ has only positive eigenvalues.
Can the requirement for $A$ to be positive semidefinite (in the sense I have defined it) be expressed as some conditions on $v$?
Your question is unclear. Usually, when $v$ is complex, the outer product $v\otimes v$ means $vv^\ast$. In this case, $A=vv^\ast$ is always positive semidefinite.
So, I suppose your $v\otimes v$ means $vv^T$. If $v=0$, surely $A=vv^T$ is positive semidefinite. Assume that $v$ is nonzero. Let $u$ and $w$ be its real and imaginary parts (i.e. $v=u+iw$). Since your question states that $vv^T$ is a projection matrix, we have $(vv^T)(vv^T)=vv^T$ and hence $v^Tv=1$. Consequently $(u+iw)^T(u+iw)=\|u\|^2-\|w\|^2+2iu^Tw=1$. Hence $\|u\|^2=\|w\|^2+1$ and $u\perp w$.
Now, if $\Re(x^{*} vv^T x) \ge 0$ for all complex vector $x$, the inequality must hold in particular for $x=w$. Yet, as $u\perp w$, $$ \Re(w^{*} vv^T w) = \Re(w^T vv^T w) = \Re((w^T(iw)(iw^T)w) = -\|w\|^4. $$ Hence $w$ must be the zero vector, i.e. $v=u$ is a real vector. Hence the necessary and sufficient condition for $A=vv^\ast$ to be positive semidefinite (given that $A$ is a projection matrix) is that $v$ is real.