Normalizing Cepstrum by Energy
Nevermind, I figured it out. For my purposes, all I really want is the cepstrum of the energy-normalized signal, which after working through the algebra is equivalent to adding a delta function to the original cepstrum. (See below.)

For a real, discrete, finite-time signal $\vec{s}$, we can compute its energy as $E(\vec{s})=\vec{s}\cdot\vec{s}$. To energy-normalize the signal (i.e. ensure that $\text{energy}=1$), we multiply $\vec{s}$ by the scalar $E(\vec{s})^{-1/2}$.

The cepstrum of the the signal $\vec{s}$ is defined as $\text{cep}(\vec{s}) = \mathfrak{F}^{-1}\big(\text{log}\big(\big|\mathfrak{F}(\vec{s})\big|^2\big)\big)$.
Therefore the cepstrum of the energy-normalized signal can be computed as: $\text{cep}\big(E(\vec{s})^{-1/2}\cdot\vec{s}\big)\\ = \mathfrak{F}^{-1}\big(\text{log}\big(\big|\mathfrak{F}\big(E(\vec{s})^{-1/2}\cdot\vec{s}\big)\big|^2\big)\big)\\ =
\mathfrak{F}^{-1}\big(\text{log}\big(\big|E(\vec{s})^{-1/2}\cdot\mathfrak{F}(\vec{s})\big|^2\big)\big)\\ = 
\mathfrak{F}^{-1}\big(\text{log}\big(E(\vec{s})^{-1}\cdot\big|\mathfrak{F}(\vec{s})\big|^2\big)\big)\\ = 
\mathfrak{F}^{-1}\big(\text{log}\big(E(\vec{s})^{-1}\big)\cdot\mathbf{1} + \text{log}\big(\big|\mathfrak{F}(\vec{s})\big|^2\big)\big)\\ = 
\mathfrak{F}^{-1}\big(\text{log}\big(\big|\mathfrak{F}(\vec{s})\big|^2\big) - \text{log}\big(E(\vec{s})\big)\cdot\mathbf{1}\big)\\ = 
\mathfrak{F}^{-1}\big(\text{log}\big(\big|\mathfrak{F}(\vec{s})\big|^2\big)\big) - \mathfrak{F}^{-1}\big(\text{log}\big(E(\vec{s})\big)\cdot\mathbf{1}\big)\\ = 
\text{cep}(\vec{s}) - \big[\text{log}\big(E(\vec{s})\big)\;\; 0\;\; 0\;\; \dots\;\; 0\big]$

Where $\mathbf{1}$ denotes the vector of all 1's.