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Let $G$ be a finite group and let $N\left(G\right)$ denote the set of conjugacy class sizes of $G$. Thompson’s conjecture states that if $G$ is a centerless group and $S$ is a non-abelian simple group satisfying $N\left(G\right)=N\left(S\right)$, then $G\cong S$. In this paper, we investigate a variation of this conjecture for some symmetric groups under a weaker assumption. In particular, it is shown that $G\cong \mathrm{Sym}(p+1)$ if and only if $\left|G\right|=(p+1)!$ and $G$ has a special conjugacy class of size $(p+1)!/p$, where $p>5$ is a prime number. Consequently, if $G$ is a centerless group with $N\left(G\right)=N\left(\mathrm{Sym}\right(p+1\left)\right)$, then $G\cong \mathrm{Sym}(p+1)$.