= Solution
On the <complex vector space> $\operatorname{CF}(G)$ of <class functions>, define the <linear operator>
$$
Tf(g)=f(g^{-1}).
$$
This is a complex-<linear map> and satisfies $T^2=I$. It is well-defined on <class functions>, because <inversion of conjugacy classes> takes an entire <conjugacy class> $K$ to another <conjugacy class> $K^{-1}$.
Use first the <basis> consisting of the <indicator functions> $1_K$ of the <conjugacy classes>. The <linear operator> $T$ permutes this <basis> by $1_K\mapsto1_{K^{-1}}$. A <permutation matrix> has <trace> equal to the number of fixed <basis vectors>, so
$$
\operatorname{tr}T=\#\{K:K=K^{-1}\}.
$$
The right-hand side counts the <real conjugacy classes>.
By <character orthogonality> and completeness, the <irreducible characters> are another <basis> of $\operatorname{CF}(G)$. Part (a) gives $T\chi=\overline\chi$. The pointwise <complex conjugate> of an <irreducible character> is again an <irreducible character>: conjugate the representing <matrices>, and conjugate any hypothetical proper <invariant subspace> back to contradict irreducibility. Thus $T$ also permutes this <basis>, and
$$
\operatorname{tr}T=\#\{\chi\in\operatorname{Irr}(G):\chi=\overline\chi\}.
$$
Here $\chi=\overline\chi$ means exactly that $\chi$ is a <real-valued character>. Equality of the <trace> in these two <bases> proves
$$
\boxed{\#\{\text{real-valued irreducible characters}\}
=\#\{\text{real conjugacy classes}\}.}
$$
This is the <real character and real conjugacy class counting theorem>. The <linear operator> used in the proof is $f\mapsto f\circ\operatorname{inv}$; on the <irreducible character> <basis> it agrees with pointwise <complex conjugation>, although pointwise <complex conjugation> itself is not a complex-<linear map> on all <class functions>.
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