It's easiest to answer this in terms of conserved densities and currents. A general local conservation law has the form $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\mathbf{J}$$
For the Schrodinger equation, multiplying through by the complex conjugate, we have $$\psi^\star i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\psi^\star\nabla^2\psi+V(x)\psi^\star\psi$$
Taking the complex conjugate of the above expression and then subtracting it from the above, we have $$\begin{align}i\hbar (\psi^\star\partial_t\psi+\psi\partial_t\psi^\star)&=-\frac{\hbar^2}{2m}(\psi^\star\nabla^2\psi+\psi\nabla^2\psi^\star)\\
\partial_t(\psi^\star\psi)&=\nabla\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)
\end{align}$$
Thus, we have a conserved quantity, $\rho=\phi^\star\phi$. This can be interpreted as the probability density of finding the particle; it is positive definite. If we do this for Klein-Gordon, though, we have $$\begin{align}(\phi^\star\partial^2_t\phi-\phi\partial^2_t\phi^\star)&=c^2(\psi^\star\nabla^2\psi-\psi\nabla^2\psi^\star)\\
\partial_t(\phi^\star\partial_t\phi-\phi\partial_t\phi^\star)&=\nabla\cdot(c^2(\psi^\star\nabla\psi-\psi\nabla\psi^\star))
\end{align}$$
Thus, the conserved quantity here is $\rho=\phi^\star\partial_t\phi-\phi\partial_t\phi^\star$; this is not always positive definite, and thus can't possibly be interpreted as probability.
Note that this lines up with Mr. Stone's answer. For Schrodinger, we have
$$
\begin{align}
\frac{d}{dt}\langle\psi|\psi\rangle &= \int_\mathcal{V} \psi^\star\psi\\
&=\int_\mathcal{V}dV\partial_t\rho\\
&=\int_\mathcal{V}dV\, \nabla\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\
&=\int_\mathcal{S}d\mathbf{a}\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\
&= 0
\end{align}
$$
because $\mathcal{V}$ is over all space. Thus, the Schrodinger norm is time-independent. For Klein-Gordon, it's the exact same business
$$
\begin{align}
\frac{d}{dt}\langle\psi|\psi\rangle &= \int_\mathcal{V} (\phi^\star\partial_t\phi-\phi\partial_t\phi^\star)\\
&=\int_\mathcal{V}dV\partial_t\rho\\
&=\int_\mathcal{V}dV\, \nabla\cdot\left(c^2(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\
&=\int_\mathcal{S}d\mathbf{a}\cdot\left(c^2(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\
&= 0
\end{align}
$$