I think the most obvious way to do this is to treat entropy as the analogue to variance, if mutual information is the analogue to covariance. Notice that one similarity is that $\mathbb{I}[X;X] = \mathbb{H}[X]$ (as for variance vs covariance) one difference is that (unlike covariance) we have $\mathbb{I}[X;Y] \geq 0 $. Then:
$$
\mathbb{I}_N[X;Y] = \frac{\mathbb{I}[X;Y]}{\sqrt{\mathbb{H}[X]\phantom{|}}\sqrt{\mathbb{H}[Y]\phantom{|}}}
$$
This is on the wikipedia page currently, among other measures. See also this question.
Also note that the following hold:
\begin{align}
\mathbb{I}[X;Y] &= \mathbb{H}[X] + \mathbb{H}[Y] - \mathbb{H}[X,Y] \\
\mathbb{H}[X] + \mathbb{H}[Y] &\geq \mathbb{H}[X,Y] \geq 0 \\
\mathbb{H}[X,Y] &= \mathbb{H}[X|Y] + \mathbb{H}[Y]
\end{align}
Since Shannon entropy and mutual information are non-negative, so is $\mathbb{I}_N[X;Y]$. However, if $X=Y$,
$$ \frac{\mathbb{H}[X] + \mathbb{H}[X] - \mathbb{H}[X,X]}{\sqrt{\mathbb{H}[X]\phantom{|}}\sqrt{\mathbb{H}[X]\phantom{|}}} = 1 $$
Another natural normalization is:
$$
\widetilde{\mathbb{I}}_N[X;Y] = \frac{\mathbb{I}[X;Y]}{{\mathbb{H}[X]}+{\mathbb{H}[Y]}}
$$