The matrix you gave has characteristic polynomial $$ p(\lambda ) = \lambda ^6 - x \lambda^5 - 0.504 \lambda + 0.504 x - 0.6048 $$ A necessary condition for $1$ to be an eigenvalue of the matrix is that $p(1) = 0$. This gives us an equation for $x$:
$$0 = 1 - x -0.504 + 0.504x - 0.6048$$ Solving for $x$ we get $$x = -0.219355 $$ Not sure if that value makes sense biologically, as all the other entries are positive, but if you want an eigenvalue of $1$ for the matrix, you'll have to go with that $x$.