Klein's modular function $J(z)$ is defined and studied in e.g. Apostol's book Modular functions and Dirichlet series in number theory.
Certain specific evaluations are available, for example,
$$J(i) = 1$$
Additionally, it is known that $J(z)$ takes on all complex values.
Question: Can one solve for that $z$ in the fundamental domain
(either explicitly or numerically) which satisfies $J(z) = \mathrm{i}$?
On
Yes, the inverse function of the $j$-invariant can be expressed in terms of the hypergeometric function. Since,
$$j(\tau) = 1728J(\tau)\tag{1}$$
we can solve for $\tau$ as,
$$\tau = i\frac{{}_2F_1(1/6, 5/6; 1; 1-\alpha)}{{}_2F_1(1/6, 5/6; 1; \alpha)}\tag{2}$$
where $\alpha$ is any root of the quadratic,
$$4\alpha(1-\alpha)=\frac{1}{J(\tau)}\tag{3}$$
Since your post is about $J(\tau) = i$, plugging that into (3), solving for $\alpha$, then plugging either root to (2), we find both,
$$\tau \approx 0.199329+0.7783729 i$$
$$\tau \approx -0.308752+1.2056644 i$$
yield
$$J(\tau) = i$$
Having numerically determined $\tau$, I then tried Mathematica's Recognize function to see if it had a closed-form, but $\tau$ does not seem to be an algebraic number of deg $< 8$.
Yes, it is possible. Here's the way to do it: choose your favorite elliptic curve $E/\mathbf C$ with $j(E) = i$. For example, the curve
$$y^2 +xy = x^3 -36(j-1728)^{-1}x - (j-1728)^{-1}$$
has $j(E)=j$ for $j \neq 0, 1728$ (so, just plug $j=i$). Compute the periods $\omega_1$ and $\omega_2$. Then $\tau = \omega_2/\omega_1$ satisfies $j(\tau)=i$.