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 $U$ in the upper half-plane (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 the tail of the cusp $U \setminus (U \cap D)$ 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 , i.e., $\tilde{D} \to the large size of D$ is surjective and generically one-to-one, but is 2-to-1 over $D$. D \cap (-\infty, 1728)$. Because$D$contains 1728,$\tilde{D}$is homeomorphic to a closed disc. We may then define$j^{-1}$as a continuous function from$\tilde{D}$to$U \cap D$by analytic continuation, and since$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$converges reasonably quickly, so it is not hard to find fixed points numerically. As far as I can tell, the fixed points don't have any particularly interesting arithmetic properties (but I would be interested to hear otherwise). Post Undeleted by S. Carnahan 2 Added sketch of proof. That said, any normalization has infinitely many fixed points. In fact, any choice of contractible I'll give an argument for the usual normalization$q^{-1} + 744 + \cdots$, first. Consider the standard fundamental domain for$U$(namely, the action of one bounded by the lines$PSL_2(\mathbb{Z})$has a fixed point in its closurei\mathbb{R} \pm 1/2$ and the unit circle), and let $\partial U$ be the boundary. You can prove this using Brouwer's fixed-point theorem together with The function $j$ takes $\partial U$ to the fact that 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.