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Let $X$ and $Y$ be two quandles and $f: X \rightarrow Y$ be a quandle homomorphism. Then we can define a map $\bar f: Inn(X) \rightarrow Inn(Y)$ as $\bar f(S_a)=S_{f(a)}$, where $a \in X$. Then $\bar f$ may not be a group homomorphism. But I am not able to construct such example.

I have posted this question on Math Stack Exchange, but have not got any reply.The link is here https://math.stackexchange.com/questions/2859327/quandle-homomorphism-does-not-always-induces-group-homomorphim-on-inner-automorp.

Any suggestions or hints would be of great help.

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    $\begingroup$ The relevant point to note is that the induced map between the inner groups becomes a group morphism whenever the quandle map is onto. If you choose a non surjective quandle map it shouldn't be difficult to construct as many examples as you like. $\endgroup$
    – Giuliano Bianco
    Commented Jul 24, 2018 at 11:04

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Let's make sure we agree on definitions.

A quandle is an algebraic structure $(A,*)$ for which each right multiplication map $S_a(x)= x*a$ by an element $a\in A$ is an automorphism fixing $a$. These right multiplications are called inner automorphisms of the quandle, and $\textit{Inn}(A)$ is the group generated by these right multiplications.

Let $G$ be a nonabelian group and define a quandle operation on $G$ by $a*b = b^{-1}ab$. Let $f\colon G\to G$ be a constant function whose range is $\{c\}$ for some noncentral element $c\in G$. The function $f$ is a quandle endomorphism of $(G,*)$, since quandles are idempotent. Also $\widetilde{f}(S_g) = S_{f(g)}=S_c$ for all $g\in G$, even for $g=1$. So, if $g=1$ we have $\textrm{id}=S_1=S_g \neq S_c$, and $\tilde{f}(\textrm{id})=\tilde{f}(S_g)=S_{f(g)}=S_c\neq \textrm{id}$.

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