For any $x\in\mathbb{R}^n$, the function $\phi(t; x)=\exp(t A)x$ solves the initial value problem
$$\dot{y}=A y,\qquad y(0)=x$$.
Notice that $\phi(t;\cdot):\mathbb{R}^n\rightarrow\mathbb{R}^n$, given by $x\mapsto\phi(t; x)=\exp(tA)x$ is defines a family of diffeomorphisms such that
- $\phi(0;x)=x$
- $\phi(t+s;x)=\phi(t;\phi(s;x))$
Notice that if $D_0\subset\mathbb{R}^n$ is a region with finite volume. Then, as time $t$, the flow $\phi$ transform $D_0$ into $D(t):=\{\phi(t;x):x\in D_0\}=:\phi(t;D_0)$.
Recall the change of variable formula for integration in $\mathbb{R}^n$:
$$\int_{G(U)}f(y)\,dx=\int_U f(G(x))|J_G(x)|\,dx$$
where $G$ is a diffeomorphism, and $J_G$ is the determinant of the Jacobian matrix, that is
$$J_G(x)=\operatorname{det}(G'(x))$$
Then, the volume $v(t)$ of $D(t)$ is calculated by
$$\begin{align}
v(t)=\int_{D(t)}\,dy&=\int_{\phi(t;D_0)}\,dy=\int_{D_0}\left|\operatorname{det}\left(\frac{\partial}{\partial y}\phi(t;y)\right)\right|\,dy\\
&=\int_{D_0}\operatorname{det}\left(\frac{\partial}{\partial y}\phi(t;y)\right)\,dy\\
&=\int_{D_0}\operatorname{det}\big(\exp(t A)\big)\,dy=\operatorname{det}\big(\exp(t A)\big) v(0)
\end{align}$$
where the absolute signs can be dropped as $\phi(0;x)=x$ and the $\phi$'s are diffeomorphisms. From this, it follows that
$$\begin{align}
v'(t)&=v(0)\frac{d}{dt}\operatorname{det}\big(\exp(t A)\big)\tag{1}\label{one}
\end{align}$$
The posting I quoted in my comment shows how to calculate the derivative of the determinant function $\operatorname{det}:\mathbb{R}^{n^2}\rightarrow\mathbb{R}$ (see Lemma there), and uses it to calculate the derivative of $W(t; y):=\operatorname{det}\Big(\frac{\partial}{\partial y}\phi(t; y)\Big)$ with respect to $t$,
where $\phi$ is a solution to the system
$$\dot{y}=f(t, y),\qquad y(0)=x$$
and shows that
$$\begin{align}
\dot{W}(t; y)=W(t; y)(\nabla_y \cdot f)(t;\phi(t; y))\tag{2}\label{two}
\end{align}$$
In this case of your problem, $f(t; y)=Ay$ and so,
$$\nabla_y\cdot f=(\partial_{y_1},\ldots,\partial_{y_n})\cdot Ay=\operatorname{trace}(A)$$
Consequently
$$\frac{d}{dt}\operatorname{det}\big(\exp(t A)\big)=\Big(\operatorname{det}\big(\exp(t A)\big)\Big) \operatorname{trace}(A)$$
Thus, \eqref{one} reduces to
$$v'(t)=v(t)\operatorname{trace}(A)$$
the conclusion to the problems follows now immediately: $v'(t)=0$ iff $\operatorname{trace}(A)=0$. (Notice that $v(t)\neq0$ (why?))
One can avoid the derivation of the formula \eqref{two} and use Jordan's canononical decomposition $A=P^{-1}J P$, where $J$ is a canonical upper triangular matrix containing eigenvalues along the diagonal. Then
$$\phi(t;x)=\exp(tA)x=P^{-1}\exp(tJ)Px$$
and so,
$$\operatorname{det}\left(\frac{\partial}{\partial x}\phi(t; x)\right)=\operatorname{det}\left(\exp(tJ)\right)$$
It is easy to check that $$\operatorname{det}\big(\exp(tJ)\big)=\exp\left(t\sum_j\lambda_j\right)=\exp(t\operatorname{trace}(A))$$ where $\lambda_j$ are the eigenvalues (including multiplicity) of $A$.
