As Kevin Buzzard pointed out in the comments, the $j$-invariant has many possible normalizations.  Automorphisms of the affine line naturally act on the target space, so any $aj+b$ with invertible $a$ is a decent replacement.

That said, any normalization has infinitely many fixed points.  I'll give an argument for the usual normalization $q^{-1} + 744 + \cdots$, first.

Consider the standard fundamental domain $U$ (namely, the one bounded by the lines $i\mathbb{R} \pm 1/2$ and the unit circle), and let $\partial U$ be the boundary.  The function $j$ takes $\partial U$ to the real ray $(-\infty,1728]$.  Because the absolute value of $j$ increases exponentially toward cusps, there is a closed disc $D$ of radius strictly greater than 1728, centered at the origin, such that $j$ takes $U \setminus (U \cap D)$ into the complement of $D$.  Let $\tilde{D}$ be the compact analytic set formed by making a branch cut of $D$ along $j(\partial U)$ - this is contractible due to the large size of $D$.  We may then define $j^{-1}$ as a continuous function from $\tilde{D}$ to $U \cap D$, and since $U \cap j(\partial U) = \emptyset$, this can be lifted to a continuous function to $\tilde{D}$ that lands in the lift of $U \cap D$.  We therefore have a continuous function from a space homeomorphic to a disc into itself, so by Brouwer's fixed-point theorem, it has a fixed point.  The image of this point in $U$ is then a fixed point for $j$.

We may do the same for any other $SL_2(\mathbb{Z})$-translate of the fundamental domain $U$, since none of them intersect the branch locus.  If we choose an alternative normalization of $j$, we can do a similar trick for any fundamental domain that does not intersect the image of its boundary.

The $q$-expansion of $j$ converges reasonably quickly, so it is not hard to find fixed points numerically.  As far as I can tell, the fixed points don't have any particularly interesting arithmetic properties (but I would be interested to hear otherwise).