Let $U=(u_{ij}), X=\text{diag}(x_1,\ldots,x_n) \in \mathbb{C}^{n\times n}$, and set $B=(b_{ij}):=a(XU^T+UX)$. We have to find $X$ such that $\tag{1} X-I=a(XU^T+UX). $ For every $i,j$ we have $ b_{ij}=a\sum_{k=1}^n(x_i\delta_{ik}u_{jk}+u_{ik}x_k\delta_{kj})=ax_i(u_{ii}+u_{ji}). $ Therefore $ (1) \iff (x_i-1)\delta_{ij}=ax_i(u_{ii}+u_{ji}) \iff [\delta_{ij}-a(u_{ii}+u_{ji})]x_i=\delta_{ij}, $ i.e. $ (1-2au_{ii})x_i=1 \quad \forall 1 \le i \le n,\ \text{ and } (u_{ii}+u_{ji})x_i=0 \quad \forall j \ne i. $
If $U$ is such that $1-2au_{ii} = 0$ for some $i$, there is no $X$ satisfying (1).
If $U$ is such that $1-2au_{ii}\ne 0$ for every $1 \le i \le n$, and $u_{ij}+u_{ii} \ne 0$ for some $j \ne i$, there is no $X$ satisfying (1).
If $U$ is such that $1-2au_{ii}\ne 0$ for every $1 \le i \le n$, and $u_{ij}+u_{ii}=0$ for every $j \ne i$, then $ X=\text{diag}[(1-2au_{11})^{-1},\ldots,(1-2au_{nn})^{-1}]. $