Suppose $X$ is a random variable such that $\operatorname*{E}(X^n) < \infty$ for any $n \in \mathbb{Z}$. We can standardize $X$ up to the second moment by letting $Y = \frac{X - \operatorname*{E}(X)}{\sqrt{\operatorname*{E}\left(X^2\right) - \operatorname*{E}(X)^2}}$ such that $\operatorname*{E}(Y) = 0$ and $\operatorname*{E}(Y^2) = 1$. How do we construct a random variable $Z$ from $X$ such that it satisfies the following properties? The right-hand side can be any constant. I picked 0's and 1's simply because they look nice.
\begin{equation} \begin{cases} \operatorname*{E}(Z) &= 0 \\ \operatorname*{E}(Z^2) &= 1 \\ \operatorname*{E}(Z^3) &= 0 \\ \operatorname*{E}(Z^4) &= 1 \\ \end{cases} \end{equation}
I would imagine we need some higher-order operations like square root or maybe logarithm, but I cannot wrap my head around it.
No, this is not always possible. First, by Jensen's inequality, $E[Z^4] = E[(Z^2)^2] \ge E[Z^2]^2$, with equality holding iff $Z$ is actually a constant. That immediately rules out $E[Z^4] = E[Z^2]$ unless $Z^2$ is a constant.
More generally, if we don't require $\mathbb{E}[Z^4] = 1$ but still want to specify some value for $\mathbb{E}[Z^3]$ and $\mathbb{E}[Z^4]$, we still can't do this in general. For example, consider the case where $X=x_1$ with probability $p$ and $X=x_2$ with probability $q=1-p$. If $Z = f(X)$, then $Z$ can only have the values $a = f(x_1)$ or $b = f(x_2)$. We have that \begin{align*} \mathbb{E}[Z] &= pa + qb \\ \mathbb{E}[Z^2] &= pa^2 + qb^2 \\ \mathbb{E}[Z^3] &= pa^3 + qb^3 \\ \mathbb{E}[Z^4] &= pa^4 + qb^4. \end{align*} If we require $E[Z] = 0$ and $E[Z^2] = 1$, that immediately requires $a=\frac{-qb}{p}$ and $b^2 = \frac{p}{q^2+qp} = \frac{p}{q}$. This specifies $a$ and $b$ up to a sign, so we have no chance to choose $a$ and $b$ to make $\mathbb{E}[Z^3]$ or $\mathbb{E}[Z^4]$ any particular value.