Let me give some motivation, which also explains how I arrived at the question. We may let the finite group $G$ act on itself by conjugation, and this makes the group ring into a $\mathbb{Z}G$-module which affords character $ \theta = \sum\_{\chi \in {\rm Irr}(G)}(\chi \overline{\chi}).$ It follows that for each irreducible character $\mu$ of $G,$ the Schur index $m\_{\mathbb{Q}}(\chi)$ divides the multiplicity of $\mu$ in $\theta$, which is easily calculated to be $\sum\_{i=1}^{k}
\mu(x\_{i}),$ where $x\_{1},x\_{2},\ldots, x\_{k}$ are representatives for all the distinct conjugacy classes of $G$ and $k = k(G)$ is the number of conjugacy classes of $G$. The multiplicity of the trivial character in $\theta$ is $k,$ and I'll say that $\theta$ is almost multiplicity free if only the trivial character occurs with multiplicity greater than one in $\theta.$


If we knew that $\theta$ was almost multiplicity free, then we could conclude that all its irreducible constituents would have Schur index one: so that leads to the question.


One easy observation is that if $G$ is a non-trivial $p$-group for some prime $p,$ then the conjugation character only has a chance to be almost multiplicity free if $G$ has nilpotence class at most $2.$ The character is clearly a multiple of the trivial character when $G$ is Abelian. Otherwise, any irreducible character of $G$ of degree divisible by $p$ is easily checked to occur with multiplicity divisible by $p$ in the conjugation character. On the other hand, if this multiplicity is zero for each non-linear irreducible character of $G,$ then the derived group $[G,G]$ is contained in the kernel of the conjugation character, which is precisely $Z(G)$, so $G$ has class $2$. 


It seems to me that it is likely to be rare for the conjugation character to be almost multiplicity free. Apart from Abelian groups and the non-Abelian groups of order $8$ I am not aware of any other groups with this property, though I have not searched very hard. I am more interested in theoretical insights, but I am also interested in empirical evidence.
(Note added later: The answer of Mark Wildon leads to the fact that any extraspecial group of order $2^{2n+1}$ has an almost multiplicity free conjugation character. In fact, it implies that a $2$-group $G$ of class $2$ has an almost multiplicity free conjugation character if and only if $G$ is extra-special. I had better give my version of the argument in full, which benefits from insights provided by Mark's answer, since there has been some confusion in the earlier comments: note that $G$ has at most $[G:G^{\prime}] + \frac{(|G|-[G:G^{\prime}])}{4}$ irreducible characters. On the other hand, each non-linear irreducible character occurs with multiplicity zero in the conjugation character, as noted above, and each non-trivial linear character occurs with multiplicity at most $1$. Since the conjugation character has degree $|G|$ and the trivial character occurs with multiplicity $k,$ we have $|G| < k + [G:G^{\prime}].$ Hence $|G| < \frac{7[G:G^{\prime}] +|G|}{4}$, which gives $|G^{\prime}| < \frac{7}{3}$, so $|G^{\prime}| = 2.$ Hence $|C\_{G}(x)| \geq \frac{|G|}{2}$ for all $x \in G.$ It follows that $\theta = \frac{|G|}{2} 1\_{G} + \frac{|Z(G)|}{2} \rho\_{G/Z(G)},$ where $\rho$ denotes the regular character. This is almost multiplicity free if and only if $Z(G) = G^{\prime}$. Now the usual argument shows that squares are central in $G,$ so that $G/Z(G)$ is elementary Abelian, and $G$ is extra-special).


Later edit: By a result of Bob Guralnick and myself (which depends on the classification of finite simple groups), it follows that no non-Abelian finite simple group ( in fact no finite group with trivial Fitting subgroup) has almost multiplicity free conjugation character. For $\langle \theta,\theta \rangle = \sum\_{i=1}^{k}|C\_{G}(x\_{i})| > |G| + k$ for $|G| >2$ . On the other hand, if $\theta$ is almost multiplicity free, then we have $\langle \theta, \theta \rangle \leq k^{2} +k.$ Hence in that case we must have $k > |G|^{\frac{1}{2}}$, whereas Guralnick and I proved that $k(G) \leq |F(G)|^{\frac{1}{2}}|G|^{\frac{1}{2}}$ for any finite group $G,$ so $k(G) \leq |G|^{\frac{1}{2}}$ whenever $F(G) = 1.$ 


(Probably last edit): Peter Mueller pointed out that the symmetric group $S\_{3}$ satisfies the condition. Let me flesh out the claim I made in comments that no other Frobenius group has an almost multiplicity free conjugation character, since F. Ladisch uses that in his definitive answer. Let $G = KH$ be a Frobenius group with kernel $K$ and complement $H.$
We recall thaat two elements of $H$ which are conjugate in $G$ are already conjugate inn $H,$ and tht every non-identity element of $G$ is conjugate either to an element of $H$ or an element of $K,$ but not both. 


Let $e = |H|.$ Then $G$ has $k(H) +\frac{|K|-1}{e}$ conjugacy classes. We note that if $\mu$ is an irreducible character of $G$ with $K$ in its kernel, then $\langle \theta, \mu \rangle$ is equal to $\mu(1)\frac{|K|-1}{e} +$ (the multiplicity of $\mu$ in the conjugation character of $H$). Since the last quantity is non-negative, we must have $|K| - 1 = e$ and $\mu(1) = 1$ (note that $\mu$ may be viewed as an irreducible character of $H).$
Thus $H$ is Abelian, and acts transitively on $K^{\#}$ by conjugation, which forces $K$ to be an elementary Abelian $p$-group for some prime $p.$ 


Now if $\chi$ is an irreducible character of $G$ which does not contain $K$ in its kernel, then $\chi$ vanishes identically outside $K$ and is induced from an irreducible character of $K,$ which we now know is linear. Thus $\chi(1) = e.$ Since $H$ acts transitively on $K^{\#},$ Clifford's Theorem tells us that $\chi(1) + e \chi(x) = 0$ for any $x \in K^{\#},$ so $\chi(x) = -1$ for each $x \in K^{\#}.$ Hence the multiplicity of $\chi$ in $\theta$ is just $e-1.$ Since $\theta$ is almost multiplicity free, we see that $e = 2$ and $|K| = 3$, as claimed.



When is the conjugation character almost multiplicity free?
Extraspecial $2$-groups and $S\_3$ are the only nonabelian examples.  

We may write $\theta$ as a sum of the permutation characters on the orbits, which in our case are the conjugacy classes: In notation of the question,
$$ \theta = \sum\_{i=1}^k (1\_{C\_G(x\_i)})^G .$$  

Suppose that $x$ and $y$ are not conjugate. If $\theta$ is almost multiplicity free, then the only common constituent of $(1\_{C\_G(x)})^G$ and $(1\_{C\_G(y)})^G$ must be $1\_G$. Thus (by Frobenius reciprocity and Mackey)  

$$ 1 = [(1\_{C\_G(x)})^G, (1\_{C\_G(y)})^G] 
 = \sum\_t [ 1\_{C\_G(x)^t\cap C\_G(y)}, 1\_{C\_G(x)^t\cap C\_G(y)}], $$
the last sum running over double coset representatives with respect to $C\_G(x)$ and $C\_G(y)$. Thus $G$ has the following property: 



> 
> **Property:** If $x$ and $y\in G$ are not conjugate, then $G= C\_G(x)C\_G(y)$. 
> 
> 
> 


The result follows from this apparently weaker property. (I didn't use that $(1\_{C\_G(x)})^G$ itself is multiplicity free.) Some first consequences: 


1. If $g\not\in Z(G)$ and $k\in \mathbb{N}$, then either $g^k\in Z(G)$ or $g$ and $g^k$ are conjugate.
2. All elements outside the center have prime power order and prime order modulo the center. (Apply 1. to prime divisors of the order of $g$.)
3. Since elements in different cosets of $G'$ are not conjugate, it follows that $G/G'$ is elementary $2$-abelian.
4. If $g\not\in Z(G)$ and $z\in Z(G)$, then $g$ and $gz$ are conjugate. In particular, $Z(G)\leq G'$, if $G$ is not abelian.


Thus if the prime $p$ divides $|Z(G)|$, then $G$ is a $p$-group (as $g$ and $gz$ have the same order by 4.), and thus by 3., $p=2$. The exponent of $G/Z(G)$ is $2$ by no.2., so $G'=Z(G)$. It is easy now to see that $|Z(G)|=2$. (See Mark Wildon's answer or the proof in the question). 


It remains to do the case $Z(G)=1$. Let $g$ be an element of maximal order $p$ (a prime, by no.2.). Let $k\in \mathbb{N}$ have multipicative order $p-1$ mod $p$. Then $g$ and $g^k$ are conjugate (1.), via $x\in G$ (say). Then $g\in C\_G(x^{p-1})$, but $g\not\in C\_G(x^n)$ for any $n< p-1$. It follows that $x^{p-1}\in Z(G)=1$ and that $p-1$ is the order of $x$. But then $p-1$ is prime and thus $p=3$. Thus $G$ is a $2$-$3$-group.  

Let $P$ be a Sylow $2$-subgroup and $Q$ a Sylow $3$-subgroup. Since all elements in $G$ have prime order, it follows that $C\_G(x)\leq P$ for all $1\neq x\in P$. Therefore, by the above "Property", all elements of $P\setminus 1$ are conjugate in $G$. The same holds for $Q\setminus 1$, too. Since $G$ is solvable (Burnside), one of the Sylows is normal and $G$ is a Frobenius group. But then $G=S\_3$ by Geoff Robinson's argument in the comments. (More directly, as $G/G'$ is a nontrivial $2$-group, we see that $Q$ is minimal normal. But then $|P|=2$ by a well-known property of Frobenius complements and since $P$ is elementary abelian, and $|Q|=3$ since $Q\setminus 1$ is a conjugacy class of $G$.)


