(1) Is wrong. You are failing to take into account there can be linear combinations of non-fixed-points which are fixed points of the conjugation action. Thus, $|Z(G)|$ is just a lower bound for the multiplicitiy of the trivial rep in the conjugation rep. Indeed, the trivial-rep isotypic component (i.e. sum of all the trivial subreps) of the conjugation rep admits a basis of the form $\{\sum_{g\in C}g\mid C\}$ as $C$ runs over the conjugacy classes of $G$. Since central elements form singleton conjugacy classes, these are the single-summand elements of this basis. But all you get is that the number of conjugacy classes equals itself.
You will basically have to learn Artin-Wedderburn if you want to understand this fact. You should already know there are algebra homomorphisms $\mathbb{C}[G]\to\mathrm{End}(V)$ for each irrep $V$. This means you can put them together into a homomorphism $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ (the sum is over irreps). This must have trivial kernel, since any element of the kernel would act as $0$ on any irrep, hence on any rep (by semisimplity), hence on the regular rep $\mathbb{C}G$ hence it times $1$ would be $0$ within $\mathbb{C}[G]$. To see that the map is also onto (and hence an algebra isomorphism), we can show the dimensions match.
To do this, let's find the multiplicity of any irrep $V$ within $\mathbb{C}G$. That equals the dimension of the hom space $\hom_G(\Bbb CG,V)$ (which you can show using semisimplicity and the distributivity of $\mathrm{hom}$). But any equivariant map $\mathbb{C}G\to V$ is determined by where $1$ is sent to, which is arbitrary, hence this hom-space is naturally isomorphic (as a vector space) to $V$ itself, so the dimension (hence the multiplicity) is $\dim V$ itself. As a result, we get $\dim\mathbb{C}[G]=\sum(\dim V)^2$.
On the one hand, the multiplicity of the trivial rep in the conjugaction action is the number of conjugacy classes. (More generally, if $\Omega$ is a $G$-set, the multiplicity of the trivial rep in the permutation rep $\mathbb{C}\Omega$ equals the number of orbits, by using a basis like the one I described for the conjugation rep. Can you show this? It is a nice exercise, and relevant to Burnside's counting lemma.) On the other hand, the trivial representation under the conjugation action corresponds to the center $Z(\mathbb{C}[G])$ of the group algebra, which has the same dimension as $\bigoplus Z(\mathrm{End}(V))$, which equals the number of irreps.
Yes, in fact we can write down an even better bound than this. If $\chi$ is a faithful character and $\psi$ is an irreducible character, then for $n \ge 1$ the sequence $\langle \chi^n, \psi \rangle$ satisfies a linear recurrence relation of order the number $d$ of distinct nonzero values that the character $\chi$ takes (which satisfies $d \le c(G)$, the number of conjugacy classes, but can be strictly less than it). (We need to take $n \ge 1$ so we can ignore the zero values.) Hence if it is zero for $n = 1, \dots d$ then it vanishes identically (since we can compute the rest of the terms using the linear recurrence). So any of the other arguments which show that this sequence is not identically zero in fact show that $\langle \chi^n, \psi \rangle \neq 0$ for some $\boxed{ 1 \le n \le d }$.
This argument can be extended slightly to give a proof, which is also given in the linked MO thread. We just consider the generating function
$$\sum_{n \ge 0} \langle \chi^n, \psi \rangle t^n = \sum_{n \ge 0} \frac{1}{|G|} \sum_{g \in G} \chi^n(g^{-1}) \psi(g) t^n = \frac{1}{|G|} \sum_{g \in G} \frac{\psi(g)}{1 - \chi(g^{-1}) t}.$$
Because $\chi$ is faithful, $g = e$ is the only element satisfying $\chi(g) = \chi(e)$, so the corresponding term $\frac{\psi(e)}{1 - \chi(e) t}$ is not canceled out by any other term in the above sum (it is the dominant singularity of the generating function), so the above generating function is nonzero and hence has some nonzero coefficient. Alternatively we can use a Vandermonde matrix, which is also alluded to in the linked MO thread.
As a simple application of this stronger bound, if we take $\chi$ to be the character of the regular representation then $d = 1$ which gives that every irreducible appears in the regular representation. Of course we knew this already but it's nice to see that this result is strong enough to give it.
As a more interesting application, let $V$ be the $n$-dimensional permutation representation of $S_n$. The character of this representation is the number of fixed points of a permutation, which takes exactly $d = n-1$ distinct nonzero values, namely $1, 2, \dots n-2, n$ (so excluding $n-1$). We conclude that every irreducible of $S_n$ appears in $V^{\otimes k}$ for some $1 \le k \le n - 1$, which is much better than either the bound $|S_n| = n!$ or the bound $|c(S_n)| = p(n)$. We ought to have a similar bound for induced representations but I don't know what that bound should be exactly.
Best Answer
Proposition: If $G$ is a finite group then there is a homomorphism $\pi:G \to \operatorname{GL}_n(\mathbb{C})$ for some positive integer $n$ so that $\chi:G \to \mathbb{C}:g \mapsto \operatorname{tr}(\pi(g))$ is injective on conjugacy classes of $G$.
I'll use the following two facts from character theory (chapter 2 and 3 of Isaacs's textbook on the Character Theory of Finite Groups):
Lemma: Two elements $g,h$ of $G$ are conjugate in $G$ if and only if $\chi(g) = \chi(h)$ for all irreducible characters $\chi$ of $G$
Lemma: $|\chi(g)| \leq \chi(1)$, $\chi(g)$ is an algebraic integer in $\mathbb{Z}[\zeta_m]$, and $\chi(1)$ is an integer dividing $m=|G|$.
Now the proof converts irreducible characters to the $N$-adic digits of a larger character.
Proof: Let $N > |G|$ and define $\chi = \theta_1 = \sum_{i=1}^k N^{i-1} \chi_i$ to be a linear combination of the $k$ irreducible characters of $G$.
If $\chi(g) = \chi(h)$ then the same is true mod $N$ in the ring of algebraic integers $\mathbb{Z}[\zeta_m]$, but then $\chi_1(g) = \chi_1(h)$ since the rest of $\chi$ is a multiple of $N$.
Now consider the remainder, $\theta_2=(\chi-\chi_1)/N = \sum_{i=2}^k N^{i-2} \chi_i$. Since both $\chi(g) = \chi(h)$ and $\chi_1(g)=\chi_1(h)$, we have $\theta_2(g)=\theta_2(h)$. Continue as before until we get $\chi_k(g)=\chi_k(h)$ and so by the lemma $g,h$ are conjugate in $G$. ∎