Sum of squares of degrees of irreducible representations whose restriction to the center is a given character equals order of inner automorphism group

Statement
Suppose $$G$$ is a finite group. Let $$Z(G)$$ denote the center of $$G$$. Suppose $$\chi$$ is a one-dimensional representation of $$G$$. Consider the subset of the set of irreducible linear representations of $$G$$ over $$\mathbb{C}$$ (up to equivalence) whose restriction to $$Z(G)$$ is an isotypical representation all of whose irreducible constituents are equal to $$\chi$$. (Note that the restriction of any irreducible linear representation to the center is isotypical as a consequence of Schur's lemma).

Then, the sum of the degrees of irreducible representations in this set equals the order of the inner automorphism group of $$G$$, i.e., it equals $$|G/Z(G)|$$.

Related facts
This result is a more fine-tuned version of the result that sum of squares of degrees of irreducible representations equals order of group, by splitting up these irreducible representations into groupings based on the restriction to the center). There will be $$|Z(G)|$$ such groupings, and the sum of squares of degrees of irreducible representations within each grouping is $$|G/Z(G)|$$.

Some cruder facts are:


 * Sum of squares of degrees of irreducible representations equals order of group
 * Order of inner automorphism group bounds square of degree of irreducible representation