If the eigenvalues of $M$ are $\lambda_1$, $\lambda_2$, ..., $\lambda_n$, then the minimum is $n (\lambda_1 \cdots \lambda_n)^{1/n}$.

**Proof:** Your problem is invariant under conjugating $X$ and $M$ by a unitary matrix. So we may assume that $M$ is diagonal, with diagonal entries $\lambda_1$, ..., $\lambda_n$. Once we have made this assumption, we have $\mathrm{Tr}(XM) = \sum x_{ii} \lambda_i$. In other words, this is a linear function of the diagonal of $X$.

We temporarily restrict ourselves to looking at those $X$ whose eigenvalues are $\mu_1$, $\mu_2$, ..., $\mu_n$. The space of positive definite matrices $X$ with eigenvalues $\mu_1$, $\mu_2$, ..., $\mu_n$ is a compact manifold. The Schur-Horn theorem states:

The subset of $\mathbb{R}^n$ which can occur as the diagonal of a positive definite matrix with eigenvalues $(\lambda_1, \ldots, \lambda_n)$ is a convex polytope;
its vertices are the $n!$ permutations
of $(\mu_1, \ldots, \mu_n)$.

A linear functional on a convex polytope is always minimized at a vertex. We conclude that

Given $(\mu_1, \ldots, \mu_n)$, as
$X$ ranges over positive definite
matrices with eigenvalues $\mu_i$, the
minimal value of $\mathrm{Tr}(XM)$ is
$\sum \lambda_i \mu_i$, where the
$\lambda$'s and the $\mu$'s are sorted
in opposite orders.

Now, we want to vary the $\mu$'s. So we want to find the minimal value of $\sum \lambda_i \mu_i$ where $\mu_i$ ranges over $n$-tuples obeying $\prod \mu_i = 1$, with $\mu_i$ sorted in the reverse order from $\lambda_i$. (In fact, removing this last condition will not effect the minimum.)

By the AM-GM inequality, $\sum \lambda_i \mu_i \geq n \prod \lambda_i^{1/n} \prod \mu_i^{1/n} = n \prod \lambda_i^{1/n}$. We get equality if $\mu_i = \left( \prod \lambda_i \right)^{1/n} / \lambda_i$, and we take $X$ to be diagonal in the same basis where $M$ is diagonal.