Trace Equivalence vs LTL Equivalence
Reading Baier and Katoen closely, they are considering both finite and infinite transition systems. See page 20 of that book for definitions.

First, take the simple transition system $EVEN$: 



Lemma: No LTL formula recognizes the language $L_{even} = $Traces$(EVEN)$. A string $c \in L_{even}$ iff $c_i = a$ for even $i$. See Wolper '81. You can prove this by first showing that no LTL formula with $n$ "next-time" operators can distinguish the strings of the form $p^i\neg p p^\omega$ for $i&gt; n$, by a simple induction.

Consider the following (infinite, non-deterministic) transition system $NOTEVEN$. Note that there are two different initial states: 



Its traces are precisely $\{a,\neg a\}^\omega - L_{even}$. 

Corollary to the Lemma: If $NOTEVEN \vDash \phi$ then $EVEN \not\vDash \neg\phi$

Now, consider this simple transition system $TOTAL$:



Its traces are clearly $\{a,\neg a\}^\omega$.

Thus, $NOTEVEN$ and $TOTAL$ are not trace equivalent. Suppose they were LTL inequivalent. Then we would have an LTL formula $\phi$ such that $NOTEVEN \vDash \phi$ and $TOTAL \not\vDash \phi$. But then, $EVEN\vDash \neg\phi$. This is a contradiction.

Thanks to Sylvain for catching a stupid bug in the first version of this answer.