I'm pretty sure the answer is yes, and that the proof will use Cox rings. However, I haven't been able to find references for all the facts about Cox rings I want, so I'll leave this as CW and hope someone else fills in the gaps.

Let $\Sigma$ and $\Sigma'$ be two fans, and $X$ and $X'$ the corresponding toric varieties, with $X \cong X'$. For simplicity, I'll assume that the unit groups of $X$ and $X'$ are both trivial (i.e. equal to $k^*$). This is equivalent to assuming that $\Sigma$ (respectively $\Sigma'$) is not contained in a hyperplane.

Let $H$ be the class group of $X$ (Weil divisors modulo principal divisors). Let $G$ be the character group $\mathrm{Hom}(H, \mathbb{G}_m)$. Cox's construction gives: A variety $Y$; an action of $G$ on $Y$; a $G$-invariant open subspace $U$; and a map $U \to X$ which makes $X$ into the categorical quotient $U/G$. This is sometimes called the *universal torsor* of $X$, although it is only a torsor when $X$ is smooth (earlier false statement corrected).

To help get you oriented, if $X = \mathbb{P}^2$, then $H = \mathbb{Z}$, $G = \mathbb{G}_m$, $Y= \mathbb{A}^3$ and $U$ is the complement of the origin in $Y$.

There are two ways to think about this construction. The one I understand well is the one from Cox's original paper. Let $\rho_1$, ..., $\rho_N$ be the rays of $\Sigma$. Let $Y = \mathbb{A}^N$. Each ray represents a class in $H$, and hence a character of $G$. Let $G$ acts on the $i$-th coordinate of $Y$ by the character of $G$ corresponding to $\rho_i$. Let $(y_1, \ldots, y_N)$ be a point of $Y$. Let $i_1$, \ldots, $i_k$ be the coordiantes for which $y_{i_j} =0$. Then $(y_1, \ldots, y_N)$ is in $U$ if and only if there is a cone of $\Sigma$ containing the rays $\rho_{i_1}$, ..., $\rho_{i_k}$. I'll cite you to Cox's paper for the map $U \to X$.

That contsruction is very explicit, but it uses the fan. There is a more abstract construction, which I don't understand as well, which works for any normal variety and does not use the torus action. Specifically, for every $h \in H$, choose a specific Weil divisor $D(h)$ representing $h$. For any $h_1$ and $h_2$, choose a rational function $u(h_1, h_2)$ with divisor $D(h_1+h_2) - D(h_1) - D(h_2)$, subject to conditions to be named later.

Define the ring
$$R := \bigoplus_{h \in H} H^0(X, \mathcal{O}(D(h)))$$
with multiplication $H^0(X, \mathcal{O}(D(h_1))) \times H^0(X, \mathcal{O}(D(h_2))) \to H^0(X, \mathcal{O}(D(h_1 + h_2)))$ given by $f_1 \times f_2 = f_1 f_2 u(h_1, h_2)$. My understanding is that (1) it is always possible to choose the $u$'s such that this is a commutative and associative ring and (2) $R$ is independent of these choices, up to nonunique isomorphism. However, you should check these claims before using them.

We set $Y = \mathrm{Spec} \ R$. This comes with an obvious action of $G$, where $G$ acts on $H^0(X, \mathcal{O}(D(h)))$ by the character $h$. I am not sure how to define $U$ or the map $U \to X$, but I think there is a way. Note that none of this uses the torus action.

Now, let $(\Sigma, X)$ and $(\Sigma', X')$ be as above, and let $(Y, G, U, U \to X)$ and $(Y', G', U', U' \to X')$ be the results of the above constructions. If I am right that the Cox construction is sufficiently natural, then there should be isomorphisms $Y \cong Y'$ and $U \cong U'$, making the obvious diagrams commute. (Note that the construction of $G$ is intrinsic, so we definitely have $G \cong G'$.)

From the first description of the Cox ring, we see that $Y \cong Y' \cong \mathbb{A}^N$, so that $\Sigma$ and $\Sigma'$ both have $N$ rays.

A priori, we have $Y \cong Y'$ as $G$-varieties, and we know that each of $Y$ and $Y'$ can be identified with $\mathbb{A}^N$ so that the $G$-action becomes linear. My next claim is that the two representations of $G$ are isomorphic; i.e., we can arrange that the composition $\mathbb{A}^n \cong Y \cong Y' \cong \mathbb{A}^N$ is linear. Proof: Find a point $y$ in $Y$, fixed by the $G$-action. There is at least one such point, because the origin is fixed. Let $\phi: Y \to Y'$ be any $G$-equivariant isomorphism. Then the $G$-action on $Y$ is isomorphic to the $G$-action on the tangent space $T_y Y$, and the $G$-action on $Y'$ is isomorphic to the $G$-action on $T_{\phi(y)} Y'$. But the differential $D \phi : T_y Y \to T_{\phi(y)} Y'$ gives a linear isomorphism between the tangent spaces. Transporting structure, we have the claim.

So, $G$ acts on $Y$ and on $Y'$ by the same characters. This means that $\Sigma$ and $\Sigma'$ have the same rays. Now, looking at the combinatorics of $Y \setminus U$ and $Y' \setminus U'$, you should be able to read off the corresponding fans.

differentequivariant line bundles on a point. (Their Chern classes are different, so there can be no equivariant isomorphism between them.) $\endgroup$ – Dave Anderson Jun 23 '11 at 15:513more comments