The local fixed point index is invariant under conjugation by a local diffeomorphism

Charts and setup. Choose charts $\phi$ at $x$ and $\psi$ at $y$ with $\phi (x)=\psi (y)=0$, admissible for $f$ and $f^{\mathrm{\prime }}$ respectively, and put $k:=\phi \circ h\circ {\psi }^{-1}$, a smooth local diffeomorphism near $0$ with $k(0)=0$ and $A:=D{k}_0$ invertible; on a neighbourhood of $0$ the identity $\widehat{f^{\mathrm{\prime }}}=k^{-1}\circ \widehat{f}\circ k$ holds, and the displacement maps $g(u):=u-\widehat{f}(u)$, $g^{\mathrm{\prime }}(u):=u-\widehat{f^{\mathrm{\prime }}}(u)$ vanish only at $u=0$ in some ball. By [F1] the values ${\mathrm{ind}}_x(f)$ and ${\mathrm{ind}}_y(f^{\mathrm{\prime }})$ are computed by the normalized sphere maps of $g$ and $g^{\mathrm{\prime }}$ at any admissible radii, so it suffices to produce one common degree for such normalized maps.

Degree of the linearized comparison map. Choose $R>0$ so that $g$ is defined and nonzero on $0<\mathrm{\mid }w\mathrm{\mid }\le R$. Choose $r>0$ so that $k$ is defined and injective on $\mathrm{\mid }u\mathrm{\mid }\le r$ and $k({\bar{B}}_r)\subseteq B_R$. Fix $0<\epsilon \le r$ and an admissible radius $0<\delta \le R$ for $g$, so that $\mathrm{\Phi }(v):=g(\delta v)\mathrm{/}\mathrm{\mid }g(\delta v)\mathrm{\mid }$ has degree ${\mathrm{ind}}_x(f)$ by [F1]. For $v\in S^{n-1}$ the point $k(\epsilon v)$ is nonzero and satisfies $\mathrm{\mid }k(\epsilon v)\mathrm{\mid }<R$, and $k(\epsilon v)=\mathrm{\mid }k(\epsilon v)\mathrm{\mid }\sigma (v)$ with $\sigma (v):=k(\epsilon v)\mathrm{/}\mathrm{\mid }k(\epsilon v)\mathrm{\mid }$. For fixed $v$ the points $k(\epsilon v)$ and $\delta \sigma (v)$ lie on one ray through $0$ and have moduli in $(0,R]$, so the radial interpolation $H_t(v):=g(((1-t)\mathrm{\mid }k(\epsilon v)\mathrm{\mid }+t\delta )\sigma (v))\mathrm{/}\mid g(((1-t)\mathrm{\mid }k(\epsilon v)\mathrm{\mid }+t\delta )\sigma (v))\mid$, $t\in [0,1]$, is a homotopy of nowhere-zero maps of $S^{n-1}$ from $v\mapsto g(k(\epsilon v))\mathrm{/}\mathrm{\mid }g(k(\epsilon v))\mathrm{\mid }$ to $\mathrm{\Phi }\circ \sigma$; hence $\mathrm{deg}(g(k(\epsilon \cdot ))\mathrm{/}\mathrm{\mid }g(k(\epsilon \cdot ))\mathrm{\mid })=\mathrm{deg}\mathrm{\Phi }\cdot \mathrm{deg}\sigma$ by [L1]. By [L2], $k(u)=Au+O(\mathrm{\mid }u{\mathrm{\mid }}^2)$ with $A$ invertible, so $t\mapsto k(tv)\mathrm{/}\mathrm{\mid }k(tv)\mathrm{\mid }$ on $0<t\le \epsilon$, extended by ${\rho }_A(v):=Av\mathrm{/}\mathrm{\mid }Av\mathrm{\mid }$ at $t=0$, is a smooth homotopy (the quotient $k(tv)\mathrm{/}t={\int }_0^{1}D{k}_{stv}vds$ extends smoothly to $t=0$) from $\sigma$ to ${\rho }_A$ through nowhere-zero maps, whence $\mathrm{deg}\sigma =\mathrm{deg}{\rho }_A=\mathrm{sign}\det A$ by [L1]. Finally the normalized map of $u\mapsto A^{-1}g(k(u))$ at radius $\epsilon$ is ${\rho }_{A^{-1}}(g(k(\epsilon \cdot ))\mathrm{/}\mathrm{\mid }g(k(\epsilon \cdot ))\mathrm{\mid })$ with ${\rho }_{A^{-1}}(w)=A^{-1}w\mathrm{/}\mathrm{\mid }{A}^{-1}w\mathrm{\mid }$, so by [L1] its degree is $\mathrm{deg}{\rho }_{A^{-1}}\cdot \mathrm{sign}\det A\cdot {\mathrm{ind}}_x(f)={\mathrm{ind}}_x(f)$, because $\mathrm{deg}{\rho }_{A^{-1}}=\mathrm{sign}\det A^{-1}=\mathrm{sign}\det A$.

Second-order comparison. Write $\widehat{f}(u)=u-g(u)$ and compute, using [L2] for $k^{-1}$ at the point $k(u)$ with increment $-g(k(u))$, $g^{\mathrm{\prime }}(u)=u-k^{-1}(k(u)-g(k(u)))=d(k^{-1}{)}_{k(u)}(g(k(u)))+O(\mathrm{\mid }g(k(u)){\mathrm{\mid }}^2)$ uniformly near $0$. Since $u\mapsto d(k^{-1}{)}_{k(u)}$ is continuous and equals the invertible $A^{-1}$ at $u=0$, on a ball of radius ${\epsilon }_0$ the estimates $\mathrm{\mid }d(k^{-1}{)}_{k(u)}v\mathrm{\mid }\asymp \mathrm{\mid }v\mathrm{\mid }$ hold uniformly, and $\mathrm{\mid }g(k(u))\mathrm{\mid }\le C\mathrm{\mid }u\mathrm{\mid }$; hence $\mathrm{\mid }{g}^{\mathrm{\prime }}(u)-A^{-1}g(k(u))\mathrm{\mid }=o(\mathrm{\mid }g(k(u))\mathrm{\mid })$ as $u\to 0$, uniformly, and in particular, after shrinking the radius $\epsilon$ fixed in step 1.2 if necessary, $\mathrm{\mid }{g}^{\mathrm{\prime }}(u)-A^{-1}g(k(u))\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }{A}^{-1}g(k(u))\mathrm{\mid }$ for $0<\mathrm{\mid }u\mathrm{\mid }\le \epsilon$.

Consequence: equal degrees. For $0<\mathrm{\mid }u\mathrm{\mid }\le \epsilon$ the vector $A^{-1}g(k(u))$ is nonzero because $g(k(u))\mathrm{\ne }0$ and $A$ is invertible, and by step 2.1 every point of the segment from $A^{-1}g(k(u))$ to $g^{\mathrm{\prime }}(u)$ lies within $\mathrm{\mid }{g}^{\mathrm{\prime }}(u)-A^{-1}g(k(u))\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }{A}^{-1}g(k(u))\mathrm{\mid }$ of $A^{-1}g(k(u))$, hence is nonzero. So $(t,u)\mapsto (1-t)A^{-1}g(k(u))+t{g}^{\mathrm{\prime }}(u)$ is a smooth homotopy of nowhere-zero maps on $S^{n-1}$, the normalized maps of $A^{-1}g(k(\epsilon \cdot ))$ and of $g^{\mathrm{\prime }}(\epsilon \cdot )$ have the same degree by [L1], and that degree is ${\mathrm{ind}}_y(f^{\mathrm{\prime }})$ by [F1].

Conclusion. Combining steps 3.1 and 1.2, the normalized maps computing ${\mathrm{ind}}_y(f^{\mathrm{\prime }})$ and ${\mathrm{ind}}_x(f)$ have the same degree, so ${\mathrm{ind}}_y(f^{\mathrm{\prime }})={\mathrm{ind}}_x(f)$. For the covering clause, $\pi$ is a local diffeomorphism, so it restricts to a diffeomorphism from an open neighbourhood of $\tilde{x}$ onto an open neighbourhood of $x$, and $\pi \circ \tilde{f}=f\circ \pi$ gives $\tilde{f}={\pi }^{-1}\circ f\circ \pi$ there; the first clause applies with $h=\pi$ and $f^{\mathrm{\prime }}=\tilde{f}$. No orientation of $M$ or $N$ is used and no choice principle is used.