Let $b\geq 2$ be an integer and $x\in[0,1[$. Let $T_b(x) = bx\mod 1$. It is known that $x$ is a normal number with respect to $b$ if and only if the sequence $(T_b^n(x))_{n\in\mathbb{N}}$ is uniformly distributed in $[0,1[$. This fact implies that if $x$ is normal with respect to $b$, then the orbit of x under $T_b$, i.e. the set
$$\operatorname{orb}(x) = \{T_b^n(x):n\in\mathbb{N}\},$$ is dense in $[0,1[$. Does the converse is true? Or do exist $b$ and $x$ such that $\operatorname{orb}(x)$ is dense, but $x$ is not normal?
The converse of this fact seems not true to me, since the sequence $(\log n\mod 1)_{n\in\mathbb{N}}$ is dense in $[0,1[$ but it's not uniformly distributed.
If every finite length block of base $b$ digits appears in the base $b$ expansion of $x$, the orbit of $x$ is dense in $[0,1[$. So construct $x$ by enumerating all finite length strings of base $b$ digits, concatenating them with interspersed "spacing" blocks of $0$, and converting the resulting digit string into the real $x$ the usual way. If you make the intervening blocks long enough, $x$ will fail to be normal. In base $b=2$, for example, start with the strings
and take the spacing blocks to be
that is, of twice the length of the preceding strings, so the concatenation is
which is has way too many $0$s in it to be normal.